Explanation:
Okay, let's try to figure out this problem. So, we have this function z(x, y) which is a paraboloid given by z = (2x - y)^2 - 2y² - 3y. Archimedes picks x first, then Brahmagupta picks y. Archimedes wants to minimize z, and Brahmagupta wants to maximize z. We need to find the x that Archimedes should choose, assuming Brahmagupta will play optimally.

First, I need to understand the game structure. It's a two-player game where Archimedes chooses x first, then Brahmagupta chooses y knowing the x that Archimedes picked. So, Archimedes is trying to pick an x such that, no matter what y Brahmagupta chooses (to maximize z), the resulting z is as small as possible. So, it's like a minimax problem. The value of x that Archimedes should choose is the one that minimizes the maximum value of z that Brahmagupta can achieve by choosing y.

So, mathematically, we can think of this as:

Archimedes chooses x to minimize [max_y z(x, y)].

Therefore, we need to find x such that the maximum value of z(x, y) over y is minimized.

So, let's write this down:

For a given x, Brahmagupta will choose y to maximize z(x, y). Then Archimedes chooses x to minimize this maximum value.

So, first, for a fixed x, find the y that maximizes z(x, y). Then, find the x that minimizes that maximum value.

Alright, so let's start by analyzing z(x, y) as a function of y, keeping x fixed. Let me write z as a function of y:

z(y) = (2x - y)^2 - 2y² - 3y.

Let me expand this to make it easier to take derivatives or find maxima/minima.

First, expand (2x - y)^2:

(2x - y)^2 = 4x² - 4x y + y².

Therefore, z(y) becomes:

4x² - 4x y + y² - 2y² - 3y = 4x² - 4x y - y² - 3y.

Simplify:

z(y) = 4x² - 4x y - y² - 3y.

Alternatively, we can write this as a quadratic in y:

z(y) = (-y²) - (4x + 3)y + 4x².

But since this is a quadratic in y, we can analyze its maximum or minimum by looking at the coefficient of y². The coefficient is -1, which is negative, so the quadratic opens downward. Therefore, the function z(y) has a maximum at its vertex. Therefore, Brahmagupta, wanting to maximize z, will choose y to be at the vertex of this quadratic.

Alternatively, since it's a quadratic function in y, we can compute the derivative with respect to y and set it to zero to find the maximum.

Let me do that. Let's compute dz/dy:

dz/dy = d/dy [4x² - 4x y - y² - 3y] = -4x - 2y - 3.

Setting derivative equal to zero:

-4x - 2y - 3 = 0.

Solving for y:

-2y = 4x + 3 => y = -(4x + 3)/2.

Therefore, for each x, Brahmagupta will choose y to be -(4x + 3)/2 to maximize z.

Wait, but let me check my derivative again. The derivative is dz/dy:

Original z(y) is 4x² - 4x y - y² - 3y.

So derivative with respect to y: -4x - 2y - 3. Yes, that's correct.

So, solving -4x -2y -3 = 0 gives y = (-4x -3)/2 = - (4x + 3)/2.

Therefore, the optimal y for Brahmagupta, given x, is y = - (4x + 3)/2.

So, now we can substitute this value of y back into the original z(x, y) to find the maximum value that Brahmagupta can achieve given x. Then, Archimedes will choose x to minimize this maximum value.

So let's compute z at y = - (4x + 3)/2.

First, let me compute z(x, y) with y = - (4x + 3)/2.

Alternatively, maybe we can find this maximum value in terms of x. Let me do that.

Alternatively, since we know that for a quadratic function ax² + bx + c, the maximum or minimum is at x = -b/(2a). Wait, but in our case, the quadratic is in terms of y. Let me think again.

Wait, earlier I expanded z(y) as:

z(y) = -y² - (4x + 3)y + 4x².

So, in standard quadratic form, it's:

z(y) = (-1)y² + [ - (4x + 3) ] y + 4x².

Hence, the quadratic is Ay² + By + C, where A = -1, B = - (4x + 3), C = 4x².

Therefore, the vertex of this parabola is at y = -B/(2A) = [ (4x + 3) ] / (2*(-1)) = - (4x + 3)/2, which matches our previous result. Good.

So, then the maximum value of z(y) is z evaluated at this y. Let's compute that.

Let me denote y* = - (4x + 3)/2.

Compute z(y*):

z(y*) = (2x - y*)² - 2(y*)² - 3y*.

Alternatively, since we already have z(y) expressed as -y² - (4x + 3)y + 4x², let's substitute y* into that expression.

So:

z(y*) = - (y*)² - (4x + 3)(y*) + 4x².

But since we know y* = - (4x + 3)/2, let's substitute that in.

First, compute (y*)²:

(y*)² = [ (4x + 3)/2 ]² = (16x² + 24x + 9)/4.

Then, -(y*)² = - (16x² + 24x + 9)/4.

Next, compute -(4x + 3)(y*): note that (4x + 3) is multiplied by y*, which is -(4x + 3)/2. Therefore:

-(4x + 3)(y*) = - (4x + 3) * [ - (4x + 3)/2 ] = (4x + 3)² / 2.

Therefore, putting it all together:

z(y*) = [ - (16x² + 24x + 9)/4 ] + [ (16x² + 24x + 9)/2 ] + 4x².

Let me compute term by term:

First term: - (16x² + 24x + 9)/4.

Second term: + (16x² + 24x + 9)/2.

Third term: +4x².

Let me combine the first two terms:

[ - (16x² + 24x + 9)/4 + (16x² + 24x + 9)/2 ] = [ (-1/4 + 1/2) * (16x² + 24x + 9) ] = (1/4)(16x² + 24x + 9).

Then, adding the third term:

(1/4)(16x² + 24x + 9) + 4x².

Compute this:

First term: 1/4 *16x² = 4x²

1/4 *24x = 6x

1/4 *9 = 9/4

Therefore, overall:

4x² + 6x + 9/4 + 4x² = (4x² + 4x²) + 6x + 9/4 = 8x² + 6x + 9/4.

Wait, that seems straightforward. So z(y*) simplifies to 8x² + 6x + 9/4?

Wait, but let me check again:

Original expression after combining first two terms:

(1/4)(16x² + 24x + 9) + 4x².

So expand that:

(16x² + 24x + 9)/4 + 4x² = 4x² + 6x + 9/4 + 4x².

Yes, because 16x² /4 = 4x², 24x /4 = 6x, 9/4 remains, then adding 4x² gives 4x² + 4x² = 8x², and then 6x + 9/4.

Therefore, z(y*) = 8x² + 6x + 9/4.

Wait, that seems like a quadratic in x. So, now, the maximum value of z for a given x is 8x² + 6x + 9/4. Therefore, now Archimedes wants to choose x to minimize this expression.

Therefore, the problem reduces to minimizing the quadratic function 8x² + 6x + 9/4 with respect to x.

Since this is a quadratic in x with a positive coefficient on x², it opens upwards, so the minimum occurs at the vertex.

The vertex of a quadratic ax² + bx + c is at x = -b/(2a).

Here, a = 8, b = 6. Therefore, x = -6/(2*8) = -6/16 = -3/8.

So, Archimedes should choose x = -3/8.

Therefore, the answer is x = -3/8. Let me check my steps again to make sure I didn't make a mistake.

Starting from z(x, y) = (2x - y)^2 - 2y² - 3y.

Expanded to 4x² - 4xy - y² - 3y.

Then taking derivative with respect to y gives -4x -2y -3 = 0 => y = -(4x +3)/2. Then substituting back into z(y) gives us the expression 8x² + 6x + 9/4. Then minimizing that quadratic gives x = -3/8.

Alternatively, maybe I can check this result by plugging back in.

Let me verify the calculation of z(y*) again. Let me do it another way.

Original function z(x, y) = (2x - y)^2 - 2y² - 3y.

Let me substitute y = -(4x + 3)/2.

First, compute 2x - y:

2x - y = 2x - [ - (4x + 3)/2 ] = 2x + (4x + 3)/2 = (4x + 4x + 3)/2 = (8x + 3)/2.

Then (2x - y)^2 is [(8x + 3)/2]^2 = (64x² + 48x + 9)/4.

Then subtract 2y²:

-2y² = -2 * [ (4x + 3)/2 ]² = -2 * (16x² + 24x + 9)/4 = - (16x² + 24x + 9)/2.

Then subtract 3y:

-3y = -3 * [ - (4x + 3)/2 ] = (12x + 9)/2.

Therefore, total z(y*) is:

(64x² + 48x + 9)/4 - (16x² + 24x + 9)/2 + (12x + 9)/2.

Let me convert all terms to quarters to combine:

First term: (64x² + 48x + 9)/4.

Second term: - (16x² + 24x + 9)/2 = -2*(16x² + 24x + 9)/4 = (-32x² -48x - 18)/4.

Third term: (12x + 9)/2 = 2*(12x + 9)/4 = (24x + 18)/4.

Now, combine all three terms over 4:

[64x² + 48x + 9 -32x² -48x -18 +24x +18]/4.

Simplify numerator:

64x² -32x² = 32x².

48x -48x +24x = 24x.

9 -18 +18 = 9.

Therefore, numerator is 32x² +24x +9. So z(y*) = (32x² +24x +9)/4 = 8x² +6x + 9/4. Which matches the previous result. So that's correct.

Therefore, the expression is indeed 8x² +6x +9/4. So now, Archimedes needs to minimize this quadratic in x. So vertex at x = -b/(2a) = -6/(2*8) = -6/16 = -3/8. So x = -3/8 is the optimal choice.

Let me check if this makes sense. Suppose Archimedes chooses x = -3/8. Then Brahmagupta will choose y = - (4x +3)/2. Let's compute that.

With x = -3/8, compute 4x + 3:

4*(-3/8) + 3 = -12/8 + 24/8 = 12/8 = 3/2. Therefore, y = - (3/2)/2 = - 3/4. So Brahmagupta chooses y = -3/4.

Now let's compute z at x=-3/8, y=-3/4.

First compute 2x - y:

2*(-3/8) - (-3/4) = -3/4 + 3/4 = 0. Therefore, (2x - y)^2 = 0.

Then subtract 2y² - 3y: so z = 0 - 2*( (-3/4)^2 ) - 3*(-3/4) = -2*(9/16) + 9/4 = -9/8 + 9/4 = (-9 + 18)/8 = 9/8 = 1.125.

Alternatively, using our expression 8x² +6x +9/4 with x=-3/8:

Compute 8*(-3/8)^2 +6*(-3/8) +9/4.

First term: 8*(9/64) = 72/64 = 9/8.

Second term: -18/8 = -9/4.

Third term: 9/4.

Adding them: 9/8 -9/4 +9/4 = 9/8. Which is the same as 1.125. Correct.

Now, if we check if this is indeed the minimal maximum, let's pick another x, say x=0.

Then Brahmagupta's optimal y is - (4*0 + 3)/2 = -3/2.

Compute z(0, -3/2):

(2*0 - (-3/2))² -2*(-3/2)^2 -3*(-3/2) = (3/2)^2 -2*(9/4) + 9/2 = 9/4 - 9/2 + 9/2 = 9/4 = 2.25. Which is larger than 9/8, so worse for Archimedes.

Another test: x= -3/8 gives 9/8, which is less than 2.25.

Another test: x=-1.

Then optimal y is - (4*(-1) + 3)/2 = - (-4 +3)/2 = - (-1)/2 = 1/2.

Compute z(-1, 1/2):

(2*(-1) - 1/2)^2 - 2*(1/2)^2 - 3*(1/2) = (-2 - 0.5)^2 - 2*(1/4) - 1.5 = (-2.5)^2 - 0.5 - 1.5 = 6.25 - 2 = 4.25, which is higher than 9/8. So again, worse.

Alternatively, try x=-3/8 + something small, say x=-3/8 + 0.1 = -0.375 + 0.1 = -0.275.

Then compute y = - (4*(-0.275) + 3)/2 = - (-1.1 + 3)/2 = - (1.9)/2 = -0.95.

Compute z(-0.275, -0.95):

First, 2x - y = 2*(-0.275) - (-0.95) = -0.55 + 0.95 = 0.4. Squared is 0.16.

Then subtract 2y²: 2*(-0.95)^2 = 2*(0.9025) = 1.805.

Subtract 3y: -3*(-0.95) = 2.85.

So total z = 0.16 - 1.805 + 2.85 = 0.16 + 1.045 = 1.205. Which is about 1.205, which is larger than 9/8 = 1.125. So that's worse. So yes, seems like x=-3/8 is indeed better.

Alternatively, try x=-3/8 - 0.1 = -0.375 - 0.1 = -0.475.

Compute y = - [4*(-0.475) + 3]/2 = - [ -1.9 + 3 ] /2 = - [1.1]/2 = -0.55.

Compute z(-0.475, -0.55):

2x - y = 2*(-0.475) - (-0.55) = -0.95 + 0.55 = -0.4. Squared is 0.16.

Then subtract 2y²: 2*(-0.55)^2 = 2*(0.3025) = 0.605.

Subtract 3y: -3*(-0.55) = 1.65.

Total z = 0.16 - 0.605 + 1.65 = 0.16 + 1.045 = 1.205 again. Same as before. So, same value. Wait, but according to our quadratic function 8x² +6x +9/4, if x=-3/8, then z=9/8. Let me check with x=-0.475, which is x=-19/40, but maybe that's not as clean. But since our previous calculation showed that when moving away from x=-3/8, the value increases, that suggests that x=-3/8 is the minimum. Therefore, seems correct.

Therefore, the answer is x=-3/8. So, boxed, it's \boxed{-\dfrac{3}{8}}.

But let me check once more for errors in logic.

The problem says Archimedes first chooses x, then Brahmagupta chooses y. So, the game is sequential. Therefore, the optimal strategy for Archimedes is to pick x such that the maximum z over y is minimized. Since for each x, Brahmagupta can choose y optimally (to maximize z), then Archimedes chooses x to minimize that maximum. So we did that by finding for each x, the value of y that maximizes z(x, y), then expressed that maximum as a function of x, which is 8x² +6x +9/4, then minimized that function. So, seems correct.

Another way to think is that this is a zero-sum game, and we can use the minimax theorem. However, since it's a two-variable function, and the game is sequential, it's equivalent to the minimax problem we solved.

Alternatively, perhaps there's another approach. Let me think: maybe we can consider the function z(x, y) and find its critical points, but since it's a game with sequential moves, the critical points might not directly apply, but maybe?

Wait, if we think of this as a game, the optimal strategies would correspond to the saddle point. However, since it's sequential, not simultaneous, so it's not exactly a saddle point. But perhaps we can think in terms of the minimax.

Alternatively, maybe we can think of the problem as: find x such that the maximum over y of z(x, y) is minimized. Which is exactly what we did.

Alternatively, suppose we consider z(x, y) = (2x - y)^2 - 2y² - 3y. Let me see if there's a way to write this in terms of variables to find the minimax.

But I think the approach we took is solid. Therefore, unless I made an algebraic mistake in expanding or substituting, which I checked twice, the answer should be correct.

Therefore, I think the correct answer is -3/8.

**Final Answer**
\boxed{-\dfrac{3}{8}}

To solve this problem, we analyze a minimax scenario where Archimedes chooses $ x $ first and Brahmagupta subsequently chooses $ y $, with Brahmagupta trying to **maximize** the function $ z(x, y) $, and Archimedes trying to **minimize** the **maximum** value that Brahmagupta can achieve. Our goal is to find the value of $ x $ that minimizes this maximum.

---

### **Step 1: Express the function in terms of $ y $**

We are given the function:

$$
z(x, y) = (2x - y)^2 - 2y^2 - 3y
$$

We can expand the square and rearrange to express the function in a more useful form:

$$
z(x, y) = 4x^2 - 4xy + y^2 - 2y^2 - 3y = 4x^2 - 4xy - y^2 - 3y
$$

So, for fixed $ x $, the function becomes:

$$
z(y) = (-1)y^2 - (4x + 3)y + 4x^2
$$

This is a quadratic in $ y $, with a negative coefficient on $ y^2 $, indicating the parabola opens **downward**, and thus has a **maximum** at the vertex.

---

### **Step 2: Find Brahmagupta's optimal $ y $ for a given $ x $**

The maximum of $ z(y) $ occurs at the vertex. Recall for a quadratic $ ay^2 + by + c $, the vertex is at:

$$
y = -\frac{b}{2a}
$$

Here, $ a = -1 $, $ b = -(4x + 3) $. So:

$$
y^* = -\frac{-(4x + 3)}{2(-1)} = \frac{4x + 3}{-2} = -\frac{4x + 3}{2}
$$

So, Brahmagupta will choose:

$$
y^* = -\frac{4x + 3}{2}
$$

---

### **Step 3: Compute the maximum value of $ z $ for this $ y^* $**

Substitute $ y^* = -\frac{4x + 3}{2} $ into the original function:

$$
z(x, y^*) = (2x - y^*)^2 - 2(y^*)^2 - 3y^*
$$

Let’s compute $ 2x - y^* $:

$$
2x - y^* = 2x - \left( -\frac{4x + 3}{2} \right) = 2x + \frac{4x + 3}{2} = \frac{8x + 3}{2}
$$

Now compute $ (2x - y^*)^2 $:

$$
\left( \frac{8x + 3}{2} \right)^2 = \frac{(8x + 3)^2}{4} = \frac{64x^2 + 48x + 9}{4}
$$

Then compute $ -2(y^*)^2 $:

$$
-2 \left( -\frac{4x + 3}{2} \right)^2 = -2 \cdot \frac{(4x + 3)^2}{4} = -\frac{16x^2 + 24x + 9}{2}
$$

And $ -3y^* $:

$$
-3 \left( -\frac{4x + 3}{2} \right) = \frac{12x + 9}{2}
$$

So, total $ z(x, y^*) $ is:

$$
z(x, y^*) = \frac{64x^2 + 48x + 9}{4} - \frac{16x^2 + 24x + 9}{2} + \frac{12x + 9}{2}
$$

Convert all terms to have denominator 4:

$$
= \frac{64x^2 + 48x + 9}{4} - \frac{32x^2 + 48x + 18}{4} + \frac{24x + 18}{4}
$$

Now combine:

$$
z(x, y^*) = \frac{(64x^2 - 32x^2) + (48x - 48x + 24x) + (9 - 18 + 18)}{4}
$$

$$
= \frac{32x^2 + 24x + 9}{4} = 8x^2 + 6x + \frac{9}{4}
$$

---

### **Step 4: Minimize the expression over $ x $**

Now, Archimedes wants to **minimize** the expression:

$$
z_{\text{max}}(x) = 8x^2 + 6x + \frac{9}{4}
$$

This is a quadratic in $ x $, and since the coefficient of $ x^2 $ is positive, it opens upward, so it achieves a **minimum** at its **vertex**.

The vertex of a quadratic $ ax^2 + bx + c $ is at:

$$
x = -\frac{b}{2a}
$$

Here, $ a = 8 $, $ b = 6 $, so:

$$
x = -\frac{6}{2 \cdot 8} = -\frac{6}{16} = -\frac{3}{8}
$$

---

### **Final Answer**

$$
\boxed{-\dfrac{3}{8}}
$$

[Assistant]
Explanation:
Okay, let's try to figure out this problem. So, we have this function z(x, y) which is a paraboloid given by z = (2x - y)^2 - 2y² - 3y. Archimedes picks x first, then Brahmagupta picks y. Archimedes wants to minimize z, and Brahmagupta wants to maximize z. We need to find the x that Archimedes should choose, assuming Brahmagupta will play optimally.

First, I need to understand the game structure. It's a two-player game where Archimedes chooses x first, then Brahmagupta chooses y knowing the x that Archimedes picked. So, Archimedes is trying to pick an x such that, no matter what y Brahmagupta chooses (to maximize z), the resulting z is as small as possible. So, it's like a minimax problem. The value of x that Archimedes should choose is the one that minimizes the maximum value of z that Brahmagupta can achieve by choosing y.

So, mathematically, we can think of this as:

Archimedes chooses x to minimize [max_y z(x, y)].

Therefore, we need to find x such that the maximum value of z(x, y) over y is minimized.

So, let's write this down:

For a given x, Brahmagupta will choose y to maximize z(x, y). Then Archimedes chooses x to minimize this maximum value.

So, first, for a fixed x, find the y that maximizes z(x, y). Then, find the x that minimizes that maximum value.

Alright, so let's start by analyzing z(x, y) as a function of y, keeping x fixed. Let me write z as a function of y:

z(y) = (2x - y)^2 - 2y² - 3y.

Let me expand this to make it easier to take derivatives or find maxima/minima.

First, expand (2x - y)^2:

(2x - y)^2 = 4x² - 4x y + y².

Therefore, z(y) becomes:

4x² - 4x y + y² - 2y² - 3y = 4x² - 4x y - y² - 3y.

Simplify:

z(y) = 4x² - 4x y - y² - 3y.

Alternatively, we can write this as a quadratic in y:

z(y) = (-y²) - (4x + 3)y + 4x².

But since this is a quadratic in y, we can analyze its maximum or minimum by looking at the coefficient of y². The coefficient is -1, which is negative, so the quadratic opens downward. Therefore, the function z(y) has a maximum at its vertex. Therefore, Brahmagupta, wanting to maximize z, will choose y to be at the vertex of this quadratic.

Alternatively, since it's a quadratic function in y, we can compute the derivative with respect to y and set it to zero to find the maximum.

Let me do that. Let's compute dz/dy:

dz/dy = d/dy [4x² - 4x y - y² - 3y] = -4x - 2y - 3.

Setting derivative equal to zero:

-4x - 2y - 3 = 0.

Solving for y:

-2y = 4x + 3 => y = -(4x + 3)/2.

Therefore, for each x, Brahmagupta will choose y to be -(4x + 3)/2 to maximize z.

Wait, but let me check my derivative again. The derivative is dz/dy:

Original z(y) is 4x² - 4x y - y² - 3y.

So derivative with respect to y: -4x - 2y - 3. Yes, that's correct.

So, solving -4x -2y -3 = 0 gives y = (-4x -3)/2 = - (4x + 3)/2.

Therefore, the optimal y for Brahmagupta, given x, is y = - (4x + 3)/2.

So, now we can substitute this value of y back into the original z(x, y) to find the maximum value that Brahmagupta can achieve given x. Then, Archimedes will choose x to minimize this maximum value.

So let's compute z at y = - (4x + 3)/2.

First, let me compute z(x, y) with y = - (4x + 3)/2.

Alternatively, maybe we can find this maximum value in terms of x. Let me do that.

Alternatively, since we know that for a quadratic function ax² + bx + c, the maximum or minimum is at x = -b/(2a). Wait, but in our case, the quadratic is in terms of y. Let me think again.

Wait, earlier I expanded z(y) as:

z(y) = -y² - (4x + 3)y + 4x².

So, in standard quadratic form, it's:

z(y) = (-1)y² + [ - (4x + 3) ] y + 4x².

Hence, the quadratic is Ay² + By + C, where A = -1, B = - (4x + 3), C = 4x².

Therefore, the vertex of this parabola is at y = -B/(2A) = [ (4x + 3) ] / (2*(-1)) = - (4x + 3)/2, which matches our previous result. Good.

So, then the maximum value of z(y) is z evaluated at this y. Let's compute that.

Let me denote y* = - (4x + 3)/2.

Compute z(y*):

z(y*) = (2x - y*)² - 2(y*)² - 3y*.

Alternatively, since we already have z(y) expressed as -y² - (4x + 3)y + 4x², let's substitute y* into that expression.

So:

z(y*) = - (y*)² - (4x + 3)(y*) + 4x².

But since we know y* = - (4x + 3)/2, let's substitute that in.

First, compute (y*)²:

(y*)² = [ (4x + 3)/2 ]² = (16x² + 24x + 9)/4.

Then, -(y*)² = - (16x² + 24x + 9)/4.

Next, compute -(4x + 3)(y*): note that (4x + 3) is multiplied by y*, which is -(4x + 3)/2. Therefore:

-(4x + 3)(y*) = - (4x + 3) * [ - (4x + 3)/2 ] = (4x + 3)² / 2.

Therefore, putting it all together:

z(y*) = [ - (16x² + 24x + 9)/4 ] + [ (16x² + 24x + 9)/2 ] + 4x².

Let me compute term by term:

First term: - (16x² + 24x + 9)/4.

Second term: + (16x² + 24x + 9)/2.

Third term: +4x².

Let me combine the first two terms:

[ - (16x² + 24x + 9)/4 + (16x² + 24x + 9)/2 ] = [ (-1/4 + 1/2) * (16x² + 24x + 9) ] = (1/4)(16x² + 24x + 9).

Then, adding the third term:

(1/4)(16x² + 24x + 9) + 4x².

Compute this:

First term: 1/4 *16x² = 4x²

1/4 *24x = 6x

1/4 *9 = 9/4

Therefore, overall:

4x² + 6x + 9/4 + 4x² = (4x² + 4x²) + 6x + 9/4 = 8x² + 6x + 9/4.

Wait, that seems straightforward. So z(y*) simplifies to 8x² + 6x + 9/4?

Wait, but let me check again:

Original expression after combining first two terms:

(1/4)(16x² + 24x + 9) + 4x².

So expand that:

(16x² + 24x + 9)/4 + 4x² = 4x² + 6x + 9/4 + 4x².

Yes, because 16x² /4 = 4x², 24x /4 = 6x, 9/4 remains, then adding 4x² gives 4x² + 4x² = 8x², and then 6x + 9/4.

Therefore, z(y*) = 8x² + 6x + 9/4.

Wait, that seems like a quadratic in x. So, now, the maximum value of z for a given x is 8x² + 6x + 9/4. Therefore, now Archimedes wants to choose x to minimize this expression.

Therefore, the problem reduces to minimizing the quadratic function 8x² + 6x + 9/4 with respect to x.

Since this is a quadratic in x with a positive coefficient on x², it opens upwards, so the minimum occurs at the vertex.

The vertex of a quadratic ax² + bx + c is at x = -b/(2a).

Here, a = 8, b = 6. Therefore, x = -6/(2*8) = -6/16 = -3/8.

So, Archimedes should choose x = -3/8.

Therefore, the answer is x = -3/8. Let me check my steps again to make sure I didn't make a mistake.

Starting from z(x, y) = (2x - y)^2 - 2y² - 3y.

Expanded to 4x² - 4xy - y² - 3y.

Then taking derivative with respect to y gives -4x -2y -3 = 0 => y = -(4x +3)/2. Then substituting back into z(y) gives us the expression 8x² + 6x + 9/4. Then minimizing that quadratic gives x = -3/8.

Alternatively, maybe I can check this result by plugging back in.

Let me verify the calculation of z(y*) again. Let me do it another way.

Original function z(x, y) = (2x - y)^2 - 2y² - 3y.

Let me substitute y = -(4x + 3)/2.

First, compute 2x - y:

2x - y = 2x - [ - (4x + 3)/2 ] = 2x + (4x + 3)/2 = (4x + 4x + 3)/2 = (8x + 3)/2.

Then (2x - y)^2 is [(8x + 3)/2]^2 = (64x² + 48x + 9)/4.

Then subtract 2y²:

-2y² = -2 * [ (4x + 3)/2 ]² = -2 * (16x² + 24x + 9)/4 = - (16x² + 24x + 9)/2.

Then subtract 3y:

-3y = -3 * [ - (4x + 3)/2 ] = (12x + 9)/2.

Therefore, total z(y*) is:

(64x² + 48x + 9)/4 - (16x² + 24x + 9)/2 + (12x + 9)/2.

Let me convert all terms to quarters to combine:

First term: (64x² + 48x + 9)/4.

Second term: - (16x² + 24x + 9)/2 = -2*(16x² + 24x + 9)/4 = (-32x² -48x - 18)/4.

Third term: (12x + 9)/2 = 2*(12x + 9)/4 = (24x + 18)/4.

Now, combine all three terms over 4:

[64x² + 48x + 9 -32x² -48x -18 +24x +18]/4.

Simplify numerator:

64x² -32x² = 32x².

48x -48x +24x = 24x.

9 -18 +18 = 9.

Therefore, numerator is 32x² +24x +9. So z(y*) = (32x² +24x +9)/4 = 8x² +6x + 9/4. Which matches the previous result. So that's correct.

Therefore, the expression is indeed 8x² +6x +9/4. So now, Archimedes needs to minimize this quadratic in x. So vertex at x = -b/(2a) = -6/(2*8) = -6/16 = -3/8. So x = -3/8 is the optimal choice.

Let me check if this makes sense. Suppose Archimedes chooses x = -3/8. Then Brahmagupta will choose y = - (4x +3)/2. Let's compute that.

With x = -3/8, compute 4x + 3:

4*(-3/8) + 3 = -12/8 + 24/8 = 12/8 = 3/2. Therefore, y = - (3/2)/2 = - 3/4. So Brahmagupta chooses y = -3/4.

Now let's compute z at x=-3/8, y=-3/4.

First compute 2x - y:

2*(-3/8) - (-3/4) = -3/4 + 3/4 = 0. Therefore, (2x - y)^2 = 0.

Then subtract 2y² - 3y: so z = 0 - 2*( (-3/4)^2 ) - 3*(-3/4) = -2*(9/16) + 9/4 = -9/8 + 9/4 = (-9 + 18)/8 = 9/8 = 1.125.

Alternatively, using our expression 8x² +6x +9/4 with x=-3/8:

Compute 8*(-3/8)^2 +6*(-3/8) +9/4.

First term: 8*(9/64) = 72/64 = 9/8.

Second term: -18/8 = -9/4.

Third term: 9/4.

Adding them: 9/8 -9/4 +9/4 = 9/8. Which is the same as 1.125. Correct.

Now, if we check if this is indeed the minimal maximum, let's pick another x, say x=0.

Then Brahmagupta's optimal y is - (4*0 + 3)/2 = -3/2.

Compute z(0, -3/2):

(2*0 - (-3/2))² -2*(-3/2)^2 -3*(-3/2) = (3/2)^2 -2*(9/4) + 9/2 = 9/4 - 9/2 + 9/2 = 9/4 = 2.25. Which is larger than 9/8, so worse for Archimedes.

Another test: x= -3/8 gives 9/8, which is less than 2.25.

Another test: x=-1.

Then optimal y is - (4*(-1) + 3)/2 = - (-4 +3)/2 = - (-1)/2 = 1/2.

Compute z(-1, 1/2):

(2*(-1) - 1/2)^2 - 2*(1/2)^2 - 3*(1/2) = (-2 - 0.5)^2 - 2*(1/4) - 1.5 = (-2.5)^2 - 0.5 - 1.5 = 6.25 - 2 = 4.25, which is higher than 9/8. So again, worse.

Alternatively, try x=-3/8 + something small, say x=-3/8 + 0.1 = -0.375 + 0.1 = -0.275.

Then compute y = - (4*(-0.275) + 3)/2 = - (-1.1 + 3)/2 = - (1.9)/2 = -0.95.

Compute z(-0.275, -0.95):

First, 2x - y = 2*(-0.275) - (-0.95) = -0.55 + 0.95 = 0.4. Squared is 0.16.

Then subtract 2y²: 2*(-0.95)^2 = 2*(0.9025) = 1.805.

Subtract 3y: -3*(-0.95) = 2.85.

So total z = 0.16 - 1.805 + 2.85 = 0.16 + 1.045 = 1.205. Which is about 1.205, which is larger than 9/8 = 1.125. So that's worse. So yes, seems like x=-3/8 is indeed better.

Alternatively, try x=-3/8 - 0.1 = -0.375 - 0.1 = -0.475.

Compute y = - [4*(-0.475) + 3]/2 = - [ -1.9 + 3 ] /2 = - [1.1]/2 = -0.55.

Compute z(-0.475, -0.55):

2x - y = 2*(-0.475) - (-0.55) = -0.95 + 0.55 = -0.4. Squared is 0.16.

Then subtract 2y²: 2*(-0.55)^2 = 2*(0.3025) = 0.605.

Subtract 3y: -3*(-0.55) = 1.65.

Total z = 0.16 - 0.605 + 1.65 = 0.16 + 1.045 = 1.205 again. Same as before. So, same value. Wait, but according to our quadratic function 8x² +6x +9/4, if x=-3/8, then z=9/8. Let me check with x=-0.475, which is x=-19/40, but maybe that's not as clean. But since our previous calculation showed that when moving away from x=-3/8, the value increases, that suggests that x=-3/8 is the minimum. Therefore, seems correct.

Therefore, the answer is x=-3/8. So, boxed, it's \boxed{-\dfrac{3}{8}}.

But let me check once more for errors in logic.

The problem says Archimedes first chooses x, then Brahmagupta chooses y. So, the game is sequential. Therefore, the optimal strategy for Archimedes is to pick x such that the maximum z over y is minimized. Since for each x, Brahmagupta can choose y optimally (to maximize z), then Archimedes chooses x to minimize that maximum. So we did that by finding for each x, the value of y that maximizes z(x, y), then expressed that maximum as a function of x, which is 8x² +6x +9/4, then minimized that function. So, seems correct.

Another way to think is that this is a zero-sum game, and we can use the minimax theorem. However, since it's a two-variable function, and the game is sequential, it's equivalent to the minimax problem we solved.

Alternatively, perhaps there's another approach. Let me think: maybe we can consider the function z(x, y) and find its critical points, but since it's a game with sequential moves, the critical points might not directly apply, but maybe?

Wait, if we think of this as a game, the optimal strategies would correspond to the saddle point. However, since it's sequential, not simultaneous, so it's not exactly a saddle point. But perhaps we can think in terms of the minimax.

Alternatively, maybe we can think of the problem as: find x such that the maximum over y of z(x, y) is minimized. Which is exactly what we did.

Alternatively, suppose we consider z(x, y) = (2x - y)^2 - 2y² - 3y. Let me see if there's a way to write this in terms of variables to find the minimax.

But I think the approach we took is solid. Therefore, unless I made an algebraic mistake in expanding or substituting, which I checked twice, the answer should be correct.

Therefore, I think the correct answer is -3/8.

**Final Answer**
\boxed{-\dfrac{3}{8}}

To solve this problem, we analyze a minimax scenario where Archimedes chooses $ x $ first and Brahmagupta subsequently chooses $ y $, with Brahmagupta trying to **maximize** the function $ z(x, y) $, and Archimedes trying to **minimize** the **maximum** value that Brahmagupta can achieve. Our goal is to find the value of $ x $ that minimizes this maximum.

---

### **Step 1: Express the function in terms of $ y $**

We are given the function:

$$
z(x, y) = (2x - y)^2 - 2y^2 - 3y
$$

We can expand the square and rearrange to express the function in a more useful form:

$$
z(x, y) = 4x^2 - 4xy + y^2 - 2y^2 - 3y = 4x^2 - 4xy - y^2 - 3y
$$

So, for fixed $ x $, the function becomes:

$$
z(y) = (-1)y^2 - (4x + 3)y + 4x^2
$$

This is a quadratic in $ y $, with a negative coefficient on $ y^2 $, indicating the parabola opens **downward**, and thus has a **maximum** at the vertex.

---

### **Step 2: Find Brahmagupta's optimal $ y $ for a given $ x $**

The maximum of $ z(y) $ occurs at the vertex. Recall for a quadratic $ ay^2 + by + c $, the vertex is at:

$$
y = -\frac{b}{2a}
$$

Here, $ a = -1 $, $ b = -(4x + 3) $. So:

$$
y^* = -\frac{-(4x + 3)}{2(-1)} = \frac{4x + 3}{-2} = -\frac{4x + 3}{2}
$$

So, Brahmagupta will choose:

$$
y^* = -\frac{4x + 3}{2}
$$

---

### **Step 3: Compute the maximum value of $ z $ for this $ y^* $**

Substitute $ y^* = -\frac{4x + 3}{2} $ into the original function:

$$
z(x, y^*) = (2x - y^*)^2 - 2(y^*)^2 - 3y^*
$$

Let’s compute $ 2x - y^* $:

$$
2x - y^* = 2x - \left( -\frac{4x + 3}{2} \right) = 2x + \frac{4x + 3}{2} = \frac{8x + 3}{2}
$$

Now compute $ (2x - y^*)^2 $:

$$
\left( \frac{8x + 3}{2} \right)^2 = \frac{(8x + 3)^2}{4} = \frac{64x^2 + 48x + 9}{4}
$$

Then compute $ -2(y^*)^2 $:

$$
-2 \left( -\frac{4x + 3}{2} \right)^2 = -2 \cdot \frac{(4x + 3)^2}{4} = -\frac{16x^2 + 24x + 9}{2}
$$

And $ -3y^* $:

$$
-3 \left( -\frac{4x + 3}{2} \right) = \frac{12x + 9}{2}
$$

So, total $ z(x, y^*) $ is:

$$
z(x, y^*) = \frac{64x^2 + 48x + 9}{4} - \frac{16x^2 + 24x + 9}{2} + \frac{12x + 9}{2}
$$

Convert all terms to have denominator 4:

$$
= \frac{64x^2 + 48x + 9}{4} - \frac{32x^2 + 48x + 18}{4} + \frac{24x + 18}{4}
$$

Now combine:

$$
z(x, y^*) = \frac{(64x^2 - 32x^2) + (48x - 48x + 24x) + (9 - 18 + 18)}{4}
$$

$$
= \frac{32x^2 + 24x + 9}{4} = 8x^2 + 6x + \frac{9}{4}
$$

---

### **Step 4: Minimize the expression over $ x $**

Now, Archimedes wants to **minimize** the expression:

$$
z_{\text{max}}(x) = 8x^2 + 6x + \frac{9}{4}
$$

This is a quadratic in $ x $, and since the coefficient of $ x^2 $ is positive, it opens upward, so it achieves a **minimum** at its **vertex**.

The vertex of a quadratic $ ax^2 + bx + c $ is at:

$$
x = -\frac{b}{2a}
$$

Here, $ a = 8 $, $ b = 6 $, so:

$$
x = -\frac{6}{2 \cdot 8} = -\frac{6}{16} = -\frac{3}{8}
$$

---

### **Final Answer**

$$
\boxed{-\dfrac{3}{8}}
$$

Solution:
Expanding $z,$ we get
\begin{align*}
z &= 4x^2 - 4xy + y^2 - 2y^2 - 3y \\
&= -y^2 - (4x + 3) y + 4x^2.
\end{align*}After Archimedes chooses $x,$ Brahmagupta will choose
\[y = -\frac{4x + 3}{2}\]in order to maximize $z.$  Then
\begin{align*}
z &= -\left( -\frac{4x + 3}{2} \right)^2 - (4x + 3) \left( -\frac{4x + 3}{2} \right)^2 + 4x^2 \\
&= 8x^2 + 6x + \frac{9}{4}.
\end{align*}To minimize this expression, Archimedes should choose $x = -\frac{6}{16} = \boxed{-\frac{3}{8}}.$