Explanation:
Okay, so I need to figure out the value of x that Archimedes should choose to minimize z, considering that Brahmagupta will choose y to maximize z after Archimedes picks x. The function given is z = (2x - y)^2 - 2y^2 - 3y.

First, let me understand the problem. It's a two-player game where Archimedes picks x first, and then Brahmagupta picks y. Archimedes wants to minimize z, and Brahmagupta wants to maximize z. So, Archimedes is trying to choose an x such that even if Brahmagupta chooses the best possible y to maximize z, the resulting z is as small as possible. This sounds like a minimax problem.

So, maybe I need to find for each x, the maximum value of z that Brahmagupta can achieve by choosing y, and then find the x that minimizes this maximum value. That makes sense.

Let me formalize this. Let me denote the function z(x, y) = (2x - y)^2 - 2y^2 - 3y.

First, for a fixed x, Brahmagupta will choose y to maximize z. So, for each x, we can consider z as a function of y, and find its maximum. Then, Archimedes will choose x to minimize that maximum value.

So, step 1: For a given x, find the value of y that maximizes z(x, y). Then, step 2: Find the x that minimizes this maximum value.

Okay, so let me start with step 1. Let me treat x as a constant and find the y that maximizes z(x, y).

First, expand the function z(x, y) to make it easier to take derivatives. Let me expand (2x - y)^2:

(2x - y)^2 = 4x² - 4xy + y².

So, substituting back into z(x, y):

z = 4x² - 4xy + y² - 2y² - 3y.

Simplify the terms:

Combine y² - 2y² = -y². So:

z = 4x² - 4xy - y² - 3y.

So, z = 4x² - 4xy - y² - 3y.

Now, since x is fixed, we can consider this as a quadratic function in y. Let me write it as:

z(y) = -y² - (4x + 3)y + 4x².

Wait, let me check that again. Let me collect the terms:

Original expression after expansion: 4x² - 4xy - y² - 3y.

So, grouping terms by y:

- y² - 4x y - 3y + 4x².

Yes, so z(y) = -y² - (4x + 3)y + 4x².

So, this is a quadratic function in y, opening downward (since the coefficient of y² is -1). Therefore, the maximum occurs at the vertex of the parabola.

For a quadratic function ay² + by + c, the vertex is at y = -b/(2a).

In this case, a = -1, b = -(4x + 3). Therefore, the vertex is at y = - [ -(4x + 3) ] / [2*(-1)] = (4x + 3)/(-2) = -(4x + 3)/2.

Wait, let me check that again.

The standard formula is y = -b/(2a). Here, the quadratic is z(y) = -y² - (4x + 3)y + 4x². So, a = -1, b = -(4x + 3). Therefore, y = -b/(2a) = - [ -(4x + 3) ] / [2*(-1)] = (4x + 3)/(-2) = -(4x + 3)/2.

Yes, that's correct. So, the maximum occurs at y = -(4x + 3)/2.

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

But wait, hold on. Since the parabola opens downward, the vertex is the maximum point. Therefore, Brahmagupta will choose this y to maximize z.

Therefore, for each x, the maximum z is achieved at y = -(4x + 3)/2.

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

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

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

So, let me compute z(x, y*).

First, let me compute (2x - y*)^2.

Compute 2x - y*:

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

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

Then, the rest of the terms: -2y*² - 3y*.

Wait, original z(x, y) is (2x - y)^2 - 2y² - 3y.

So, substituting y*:

z = (2x - y*)² - 2(y*)² - 3y*.

So, let me compute each term:

First term: (2x - y*)² = (8x + 3)² / 4.

Second term: -2(y*)². Let me compute y*:

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

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

Therefore, combining all three terms:

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

Now, let me compute this expression. Let me first compute each term separately.

First term: (8x + 3)^2 / 4. Let me expand (8x + 3)^2:

(8x)^2 + 2*8x*3 + 3^2 = 64x² + 48x + 9. Therefore, divided by 4: (64x² + 48x + 9)/4.

Second term: -(4x + 3)^2 / 2. Let me expand (4x + 3)^2:

(4x)^2 + 2*4x*3 + 3^2 = 16x² + 24x + 9. Divided by 2: (16x² + 24x + 9)/2. But with a negative sign: - (16x² + 24x + 9)/2.

Third term: 3*(4x + 3)/2 = (12x + 9)/2.

So, combining all three terms:

First term: (64x² + 48x + 9)/4

Second term: - (16x² + 24x + 9)/2

Third term: (12x + 9)/2

Let me convert all terms to have denominator 4 to combine them:

First term remains as is: (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, adding all three terms:

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

Let me compute numerator step by step:

Start with 64x² - 32x² = 32x²

Then, 48x - 48x + 24x = 24x

Then, 9 - 18 + 18 = 9

Therefore, numerator is 32x² + 24x + 9.

Therefore, z = (32x² + 24x + 9)/4.

Simplify this expression:

Divide each term by 4:

z = 8x² + 6x + 9/4.

Wait, 32x² / 4 = 8x², 24x / 4 = 6x, 9/4 remains.

So, z = 8x² + 6x + 9/4.

Wait, so after Brahmagupta chooses y optimally, the value of z is 8x² + 6x + 9/4.

Therefore, Archimedes wants to choose x to minimize this expression.

So, now, the problem reduces to minimizing the quadratic function 8x² + 6x + 9/4.

Since this is a quadratic in x with a positive leading coefficient (8), 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.

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

Wait, but let me check my calculations again to make sure I didn't make any mistakes.

Starting from z = (32x² + 24x + 9)/4.

Yes, that was after combining all terms. Let me check the numerator again:

Original terms:

First term: 64x² + 48x + 9

Second term: -32x² - 48x - 18

Third term: +24x + 18

Adding them together:

64x² - 32x² = 32x²

48x - 48x + 24x = 24x

9 - 18 + 18 = 9.

Yes, that's correct. So numerator is 32x² + 24x + 9. Divided by 4 gives 8x² + 6x + 9/4.

So, the quadratic is 8x² + 6x + 9/4.

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

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

But let me verify this result by plugging back into the original function.

Alternatively, maybe I can check if there's an error in the earlier steps.

Let me check the calculation of z at y* again.

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

We found y* = -(4x + 3)/2.

Let me compute 2x - y*:

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

So, (2x - y*)² = (8x + 3)^2 / 4.

Then, -2y*² - 3y*:

First, y* = -(4x + 3)/2.

So, y*² = (4x + 3)^2 / 4.

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

Then, -3y* = -3*(-(4x + 3)/2) = 3*(4x + 3)/2.

Therefore, total z = [(8x + 3)^2 / 4] - [(4x + 3)^2 / 2] + [3*(4x + 3)/2].

Yes, that's correct.

Then expanding:

(8x + 3)^2 = 64x² + 48x + 9. Divided by 4: 16x² + 12x + 9/4.

Wait, hold on, earlier I had written that as (64x² + 48x + 9)/4, which is 16x² + 12x + 9/4.

Then, -(4x + 3)^2 / 2: -(16x² + 24x + 9)/2 = -8x² -12x - 9/2.

Then, 3*(4x + 3)/2 = 6x + 9/2.

So adding all three terms:

First term: 16x² + 12x + 9/4

Second term: -8x² -12x - 9/2

Third term: +6x + 9/2

Adding them together:

16x² -8x² = 8x²

12x -12x +6x = 6x

9/4 -9/2 +9/2 = 9/4.

So total z = 8x² + 6x + 9/4.

Yes, same result. So that's correct.

Therefore, the expression is correct.

Therefore, the minimum of 8x² + 6x + 9/4 occurs at x = -b/(2a) = -6/(16) = -3/8.

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

But let me check if there's another way to approach this problem, maybe by considering the original function.

Alternatively, since Archimedes chooses x first, and then Brahmagupta chooses y, maybe we can think of this as a two-player game where the value is the minimax value.

Alternatively, maybe we can use calculus to find the optimal strategies.

But since we already did the calculation and arrived at x = -3/8, let me check if that makes sense.

Alternatively, let me check with specific values.

Suppose Archimedes chooses x = -3/8. Then, Brahmagupta will choose y = -(4x + 3)/2.

Let me compute y* when x = -3/8:

y* = -(4*(-3/8) + 3)/2 = -(-12/8 + 3)/2 = -(-3/2 + 3)/2 = -((3/2))/2 = -3/4.

So, y* = -3/4.

Then, compute z at x = -3/8, y = -3/4.

Original z = (2x - y)^2 - 2y² - 3y.

Compute 2x - y: 2*(-3/8) - (-3/4) = -3/4 + 3/4 = 0.

Therefore, (2x - y)^2 = 0.

Then, -2y² - 3y = -2*(9/16) - 3*(-3/4) = -9/8 + 9/4 = (-9/8 + 18/8) = 9/8.

Therefore, z = 0 + 9/8 = 9/8.

So, when x = -3/8, the maximum z Brahmagupta can achieve is 9/8.

Now, let me check if this is indeed the minimum possible.

Suppose Archimedes chooses another x, say x = 0. Then, Brahmagupta will choose y = -(4*0 + 3)/2 = -3/2.

Compute z at x = 0, y = -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.

Which is 2.25, which is larger than 9/8 (which is 1.125). So, that's worse for Archimedes.

Another test: suppose Archimedes chooses x = -3/8, which gives z = 9/8.

What if Archimedes chooses x = -1/2? Let's see.

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

Compute z at x = -1/2, y = -1/2:

(2*(-1/2) - (-1/2))² - 2*(-1/2)^2 - 3*(-1/2)

Compute 2*(-1/2) = -1, so -1 - (-1/2) = -1 + 1/2 = -1/2. Squared is 1/4.

Then, -2*(1/4) - 3*(-1/2) = -1/2 + 3/2 = 1.

So total z = 1/4 + 1 = 5/4 = 1.25, which is still higher than 9/8 = 1.125.

Another test: x = -3/8.

Alternatively, let me check if the quadratic 8x² + 6x + 9/4 indeed has its minimum at x = -3/8.

Take derivative: d/dx [8x² + 6x + 9/4] = 16x + 6. Setting to zero: 16x + 6 = 0 => x = -6/16 = -3/8. Correct.

So, that seems consistent.

Alternatively, maybe I can check another x. Let me try x = -3/8 + h, and see if the value increases.

But maybe that's overkill.

Alternatively, let me check if there's a possibility that Brahmagupta might not choose the critical point? But since the function is quadratic in y, and the coefficient of y² is negative, the maximum is indeed at the vertex. So Brahmagupta will choose that y.

Therefore, the calculation seems correct.

Therefore, the answer is x = -3/8.

But let me check once again if I didn't make a mistake in the sign when calculating y*.

Original y* was -(4x + 3)/2.

Yes, because for the quadratic z(y) = -y² - (4x + 3)y + 4x², the vertex is at y = -b/(2a) where a = -1, b = -(4x + 3). Therefore, y = - [ -(4x + 3) ] / [2*(-1)] = (4x + 3)/(-2) = -(4x + 3)/2. Correct.

Therefore, that is correct.

Therefore, I think my answer is correct.

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

To solve the problem, we are given a function:

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

This represents a paraboloid in the $xy$-plane. The game is structured as follows:

- **Archimedes** chooses $x$ first.
- **Brahmagupta** then chooses $y$ to **maximize** $z$.
- Archimedes wants to **minimize** the resulting $z$, assuming Brahmagupta plays optimally.

This is a classic **minimax problem**, where Archimedes seeks to minimize the **maximum** value of $z$ that Brahmagupta can achieve for a given $x$.

---

### Step 1: Express $z$ as a function of $y$ for fixed $x$

We expand and simplify the expression:

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

Expanding the square:

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

So:

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

Grouping terms:

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

This is a **quadratic in $y$** with a **negative leading coefficient**, so it opens **downward**, and the **maximum** occurs at the **vertex**.

---

### Step 2: Find the value of $y$ that maximizes $z$ for fixed $x$

For a quadratic $ay^2 + by + c$, the vertex occurs 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: Substitute $y^*$ back into $z(x, y)$ to find the maximum value of $z$ for a given $x$

We now compute:

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

First, compute $2x - y^*$:

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

So:

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

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

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

Then compute $-3y^*$:

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

Now combine all terms:

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

To simplify, convert all terms to have denominator 4:

- First term: $\frac{(8x + 3)^2}{4}$
- Second term: $-\frac{(4x + 3)^2}{2} = -\frac{2(4x + 3)^2}{4}$
- Third term: $\frac{3(4x + 3)}{2} = \frac{6(4x + 3)}{4}$

Now combine:

$$
z(x, y^*) = \frac{(8x + 3)^2 - 2(4x + 3)^2 + 6(4x + 3)}{4}
$$

Expand each term:

- $(8x + 3)^2 = 64x^2 + 48x + 9$
- $2(4x + 3)^2 = 2(16x^2 + 24x + 9) = 32x^2 + 48x + 18$
- $6(4x + 3) = 24x + 18$

Now combine:

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

Simplify numerator:

- $64x^2 - 32x^2 = 32x^2$
- $48x - 48x + 24x = 24x$
- $9 - 18 + 18 = 9$

So:

$$
z(x, y^*) = \frac{32x^2 + 24x + 9}{4} = 8x^2 + 6x + \frac{9}{4}
$$

---

### Step 4: Minimize the expression $8x^2 + 6x + \frac{9}{4}$

This is a quadratic in $x$ with a positive leading coefficient, so it opens **upward**, and the **minimum** occurs at the **vertex**.

The vertex of $ax^2 + bx + c$ 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, so I need to figure out the value of x that Archimedes should choose to minimize z, considering that Brahmagupta will choose y to maximize z after Archimedes picks x. The function given is z = (2x - y)^2 - 2y^2 - 3y.

First, let me understand the problem. It's a two-player game where Archimedes picks x first, and then Brahmagupta picks y. Archimedes wants to minimize z, and Brahmagupta wants to maximize z. So, Archimedes is trying to choose an x such that even if Brahmagupta chooses the best possible y to maximize z, the resulting z is as small as possible. This sounds like a minimax problem.

So, maybe I need to find for each x, the maximum value of z that Brahmagupta can achieve by choosing y, and then find the x that minimizes this maximum value. That makes sense.

Let me formalize this. Let me denote the function z(x, y) = (2x - y)^2 - 2y^2 - 3y.

First, for a fixed x, Brahmagupta will choose y to maximize z. So, for each x, we can consider z as a function of y, and find its maximum. Then, Archimedes will choose x to minimize that maximum value.

So, step 1: For a given x, find the value of y that maximizes z(x, y). Then, step 2: Find the x that minimizes this maximum value.

Okay, so let me start with step 1. Let me treat x as a constant and find the y that maximizes z(x, y).

First, expand the function z(x, y) to make it easier to take derivatives. Let me expand (2x - y)^2:

(2x - y)^2 = 4x² - 4xy + y².

So, substituting back into z(x, y):

z = 4x² - 4xy + y² - 2y² - 3y.

Simplify the terms:

Combine y² - 2y² = -y². So:

z = 4x² - 4xy - y² - 3y.

So, z = 4x² - 4xy - y² - 3y.

Now, since x is fixed, we can consider this as a quadratic function in y. Let me write it as:

z(y) = -y² - (4x + 3)y + 4x².

Wait, let me check that again. Let me collect the terms:

Original expression after expansion: 4x² - 4xy - y² - 3y.

So, grouping terms by y:

- y² - 4x y - 3y + 4x².

Yes, so z(y) = -y² - (4x + 3)y + 4x².

So, this is a quadratic function in y, opening downward (since the coefficient of y² is -1). Therefore, the maximum occurs at the vertex of the parabola.

For a quadratic function ay² + by + c, the vertex is at y = -b/(2a).

In this case, a = -1, b = -(4x + 3). Therefore, the vertex is at y = - [ -(4x + 3) ] / [2*(-1)] = (4x + 3)/(-2) = -(4x + 3)/2.

Wait, let me check that again.

The standard formula is y = -b/(2a). Here, the quadratic is z(y) = -y² - (4x + 3)y + 4x². So, a = -1, b = -(4x + 3). Therefore, y = -b/(2a) = - [ -(4x + 3) ] / [2*(-1)] = (4x + 3)/(-2) = -(4x + 3)/2.

Yes, that's correct. So, the maximum occurs at y = -(4x + 3)/2.

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

But wait, hold on. Since the parabola opens downward, the vertex is the maximum point. Therefore, Brahmagupta will choose this y to maximize z.

Therefore, for each x, the maximum z is achieved at y = -(4x + 3)/2.

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

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

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

So, let me compute z(x, y*).

First, let me compute (2x - y*)^2.

Compute 2x - y*:

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

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

Then, the rest of the terms: -2y*² - 3y*.

Wait, original z(x, y) is (2x - y)^2 - 2y² - 3y.

So, substituting y*:

z = (2x - y*)² - 2(y*)² - 3y*.

So, let me compute each term:

First term: (2x - y*)² = (8x + 3)² / 4.

Second term: -2(y*)². Let me compute y*:

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

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

Therefore, combining all three terms:

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

Now, let me compute this expression. Let me first compute each term separately.

First term: (8x + 3)^2 / 4. Let me expand (8x + 3)^2:

(8x)^2 + 2*8x*3 + 3^2 = 64x² + 48x + 9. Therefore, divided by 4: (64x² + 48x + 9)/4.

Second term: -(4x + 3)^2 / 2. Let me expand (4x + 3)^2:

(4x)^2 + 2*4x*3 + 3^2 = 16x² + 24x + 9. Divided by 2: (16x² + 24x + 9)/2. But with a negative sign: - (16x² + 24x + 9)/2.

Third term: 3*(4x + 3)/2 = (12x + 9)/2.

So, combining all three terms:

First term: (64x² + 48x + 9)/4

Second term: - (16x² + 24x + 9)/2

Third term: (12x + 9)/2

Let me convert all terms to have denominator 4 to combine them:

First term remains as is: (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, adding all three terms:

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

Let me compute numerator step by step:

Start with 64x² - 32x² = 32x²

Then, 48x - 48x + 24x = 24x

Then, 9 - 18 + 18 = 9

Therefore, numerator is 32x² + 24x + 9.

Therefore, z = (32x² + 24x + 9)/4.

Simplify this expression:

Divide each term by 4:

z = 8x² + 6x + 9/4.

Wait, 32x² / 4 = 8x², 24x / 4 = 6x, 9/4 remains.

So, z = 8x² + 6x + 9/4.

Wait, so after Brahmagupta chooses y optimally, the value of z is 8x² + 6x + 9/4.

Therefore, Archimedes wants to choose x to minimize this expression.

So, now, the problem reduces to minimizing the quadratic function 8x² + 6x + 9/4.

Since this is a quadratic in x with a positive leading coefficient (8), 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.

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

Wait, but let me check my calculations again to make sure I didn't make any mistakes.

Starting from z = (32x² + 24x + 9)/4.

Yes, that was after combining all terms. Let me check the numerator again:

Original terms:

First term: 64x² + 48x + 9

Second term: -32x² - 48x - 18

Third term: +24x + 18

Adding them together:

64x² - 32x² = 32x²

48x - 48x + 24x = 24x

9 - 18 + 18 = 9.

Yes, that's correct. So numerator is 32x² + 24x + 9. Divided by 4 gives 8x² + 6x + 9/4.

So, the quadratic is 8x² + 6x + 9/4.

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

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

But let me verify this result by plugging back into the original function.

Alternatively, maybe I can check if there's an error in the earlier steps.

Let me check the calculation of z at y* again.

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

We found y* = -(4x + 3)/2.

Let me compute 2x - y*:

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

So, (2x - y*)² = (8x + 3)^2 / 4.

Then, -2y*² - 3y*:

First, y* = -(4x + 3)/2.

So, y*² = (4x + 3)^2 / 4.

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

Then, -3y* = -3*(-(4x + 3)/2) = 3*(4x + 3)/2.

Therefore, total z = [(8x + 3)^2 / 4] - [(4x + 3)^2 / 2] + [3*(4x + 3)/2].

Yes, that's correct.

Then expanding:

(8x + 3)^2 = 64x² + 48x + 9. Divided by 4: 16x² + 12x + 9/4.

Wait, hold on, earlier I had written that as (64x² + 48x + 9)/4, which is 16x² + 12x + 9/4.

Then, -(4x + 3)^2 / 2: -(16x² + 24x + 9)/2 = -8x² -12x - 9/2.

Then, 3*(4x + 3)/2 = 6x + 9/2.

So adding all three terms:

First term: 16x² + 12x + 9/4

Second term: -8x² -12x - 9/2

Third term: +6x + 9/2

Adding them together:

16x² -8x² = 8x²

12x -12x +6x = 6x

9/4 -9/2 +9/2 = 9/4.

So total z = 8x² + 6x + 9/4.

Yes, same result. So that's correct.

Therefore, the expression is correct.

Therefore, the minimum of 8x² + 6x + 9/4 occurs at x = -b/(2a) = -6/(16) = -3/8.

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

But let me check if there's another way to approach this problem, maybe by considering the original function.

Alternatively, since Archimedes chooses x first, and then Brahmagupta chooses y, maybe we can think of this as a two-player game where the value is the minimax value.

Alternatively, maybe we can use calculus to find the optimal strategies.

But since we already did the calculation and arrived at x = -3/8, let me check if that makes sense.

Alternatively, let me check with specific values.

Suppose Archimedes chooses x = -3/8. Then, Brahmagupta will choose y = -(4x + 3)/2.

Let me compute y* when x = -3/8:

y* = -(4*(-3/8) + 3)/2 = -(-12/8 + 3)/2 = -(-3/2 + 3)/2 = -((3/2))/2 = -3/4.

So, y* = -3/4.

Then, compute z at x = -3/8, y = -3/4.

Original z = (2x - y)^2 - 2y² - 3y.

Compute 2x - y: 2*(-3/8) - (-3/4) = -3/4 + 3/4 = 0.

Therefore, (2x - y)^2 = 0.

Then, -2y² - 3y = -2*(9/16) - 3*(-3/4) = -9/8 + 9/4 = (-9/8 + 18/8) = 9/8.

Therefore, z = 0 + 9/8 = 9/8.

So, when x = -3/8, the maximum z Brahmagupta can achieve is 9/8.

Now, let me check if this is indeed the minimum possible.

Suppose Archimedes chooses another x, say x = 0. Then, Brahmagupta will choose y = -(4*0 + 3)/2 = -3/2.

Compute z at x = 0, y = -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.

Which is 2.25, which is larger than 9/8 (which is 1.125). So, that's worse for Archimedes.

Another test: suppose Archimedes chooses x = -3/8, which gives z = 9/8.

What if Archimedes chooses x = -1/2? Let's see.

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

Compute z at x = -1/2, y = -1/2:

(2*(-1/2) - (-1/2))² - 2*(-1/2)^2 - 3*(-1/2)

Compute 2*(-1/2) = -1, so -1 - (-1/2) = -1 + 1/2 = -1/2. Squared is 1/4.

Then, -2*(1/4) - 3*(-1/2) = -1/2 + 3/2 = 1.

So total z = 1/4 + 1 = 5/4 = 1.25, which is still higher than 9/8 = 1.125.

Another test: x = -3/8.

Alternatively, let me check if the quadratic 8x² + 6x + 9/4 indeed has its minimum at x = -3/8.

Take derivative: d/dx [8x² + 6x + 9/4] = 16x + 6. Setting to zero: 16x + 6 = 0 => x = -6/16 = -3/8. Correct.

So, that seems consistent.

Alternatively, maybe I can check another x. Let me try x = -3/8 + h, and see if the value increases.

But maybe that's overkill.

Alternatively, let me check if there's a possibility that Brahmagupta might not choose the critical point? But since the function is quadratic in y, and the coefficient of y² is negative, the maximum is indeed at the vertex. So Brahmagupta will choose that y.

Therefore, the calculation seems correct.

Therefore, the answer is x = -3/8.

But let me check once again if I didn't make a mistake in the sign when calculating y*.

Original y* was -(4x + 3)/2.

Yes, because for the quadratic z(y) = -y² - (4x + 3)y + 4x², the vertex is at y = -b/(2a) where a = -1, b = -(4x + 3). Therefore, y = - [ -(4x + 3) ] / [2*(-1)] = (4x + 3)/(-2) = -(4x + 3)/2. Correct.

Therefore, that is correct.

Therefore, I think my answer is correct.

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

To solve the problem, we are given a function:

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

This represents a paraboloid in the $xy$-plane. The game is structured as follows:

- **Archimedes** chooses $x$ first.
- **Brahmagupta** then chooses $y$ to **maximize** $z$.
- Archimedes wants to **minimize** the resulting $z$, assuming Brahmagupta plays optimally.

This is a classic **minimax problem**, where Archimedes seeks to minimize the **maximum** value of $z$ that Brahmagupta can achieve for a given $x$.

---

### Step 1: Express $z$ as a function of $y$ for fixed $x$

We expand and simplify the expression:

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

Expanding the square:

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

So:

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

Grouping terms:

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

This is a **quadratic in $y$** with a **negative leading coefficient**, so it opens **downward**, and the **maximum** occurs at the **vertex**.

---

### Step 2: Find the value of $y$ that maximizes $z$ for fixed $x$

For a quadratic $ay^2 + by + c$, the vertex occurs 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: Substitute $y^*$ back into $z(x, y)$ to find the maximum value of $z$ for a given $x$

We now compute:

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

First, compute $2x - y^*$:

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

So:

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

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

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

Then compute $-3y^*$:

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

Now combine all terms:

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

To simplify, convert all terms to have denominator 4:

- First term: $\frac{(8x + 3)^2}{4}$
- Second term: $-\frac{(4x + 3)^2}{2} = -\frac{2(4x + 3)^2}{4}$
- Third term: $\frac{3(4x + 3)}{2} = \frac{6(4x + 3)}{4}$

Now combine:

$$
z(x, y^*) = \frac{(8x + 3)^2 - 2(4x + 3)^2 + 6(4x + 3)}{4}
$$

Expand each term:

- $(8x + 3)^2 = 64x^2 + 48x + 9$
- $2(4x + 3)^2 = 2(16x^2 + 24x + 9) = 32x^2 + 48x + 18$
- $6(4x + 3) = 24x + 18$

Now combine:

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

Simplify numerator:

- $64x^2 - 32x^2 = 32x^2$
- $48x - 48x + 24x = 24x$
- $9 - 18 + 18 = 9$

So:

$$
z(x, y^*) = \frac{32x^2 + 24x + 9}{4} = 8x^2 + 6x + \frac{9}{4}
$$

---

### Step 4: Minimize the expression $8x^2 + 6x + \frac{9}{4}$

This is a quadratic in $x$ with a positive leading coefficient, so it opens **upward**, and the **minimum** occurs at the **vertex**.

The vertex of $ax^2 + bx + c$ 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 $x = -\frac{6}{16} = \boxed{-\frac{3}{8}}.$