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 because this seems straightforward, but I want to make sure I didn't make a mistake in the earlier steps.

Let me recap:

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

For fixed x, found y* = -(4x + 3)/2.

Substituted back into z(x, y) and after simplifying, got z = 8x² + 6x + 9/4.

Then, minimized that quadratic, getting x = -3/8.

But let me verify the substitution step again because that's where errors can creep in.

So, let me recompute z at y*:

First term: (2x - y*)².

Earlier, I found 2x - y* = (8x + 3)/2.

So, squared is (8x + 3)^2 / 4.

Second term: -2y*².

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

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

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

So, adding all three terms:

(8x + 3)^2 / 4 - (4x + 3)^2 / 2 + 3*(4x + 3)/2.

Let me compute each term again:

First term: (8x + 3)^2 / 4 = (64x² + 48x + 9)/4.

Second term: -(4x + 3)^2 / 2 = -(16x² + 24x + 9)/2.

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

Now, converting all to quarters:

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.

Adding them together:

Numerator: 64x² + 48x + 9 - 32x² - 48x - 18 + 24x + 18.

Compute term by term:

64x² - 32x² = 32x².

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

9 - 18 + 18 = 9.

So numerator is 32x² + 24x + 9. Divided by 4: (32x² + 24x + 9)/4 = 8x² + 6x + 9/4.

Yes, that's correct.

So, the expression is indeed 8x² + 6x + 9/4.

Therefore, the minimum occurs at x = -b/(2a) = -6/(2*8) = -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 game theory aspect.

Alternatively, since Archimedes chooses x first, and Brahmagupta chooses y second, we can think of this as a two-player game where the value is z(x, y). The minimax value is the minimum over x of the maximum over y of z(x, y).

So, the value we found, 8x² + 6x + 9/4, is the maximum over y of z(x, y) for each x. Then, Archimedes minimizes this.

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

But since we already did that, and the result seems consistent, maybe that's okay.

But let me check with another approach. Let me think about the original function.

Alternatively, maybe we can consider the entire game as a function of x and y, and find the saddle point.

But since Archimedes chooses x first, and then Brahmagupta chooses y, the optimal strategy for Archimedes is to choose x such that the maximum over y of z(x, y) is minimized.

Alternatively, maybe we can think of this as a zero-sum game, and find the value of the game.

But perhaps my initial approach is correct.

But let me check with specific values. Let me pick x = -3/8 and see what y Brahmagupta would choose, and compute z.

So, x = -3/8. Then, y* = -(4x + 3)/2.

Compute 4x + 3: 4*(-3/8) + 3 = -12/8 + 3 = -3/2 + 3 = 3/2.

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

So, y = -3/4.

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

First, 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.

Now, let me check if this is indeed the minimum. Let me pick another x, say x = 0.

For x = 0, y* = -(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.

Which is 2.25, while 9/8 is 1.125, so indeed lower.

Another test: x = -3/8, z = 9/8.

What if x = -1/2? Let's compute.

x = -1/2.

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

Compute z(-1/2, -1/2):

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

= (-1 + 1/2)^2 - 2*(1/4) + 3/2

= (-1/2)^2 - 1/2 + 3/2

= 1/4 - 1/2 + 3/2 = 1/4 + 1 = 5/4 = 1.25.

Which is 1.25, which is higher than 9/8 = 1.125.

So, that seems consistent.

Another test: x = -3/8.

Alternatively, let me check the expression 8x² + 6x + 9/4 at x = -3/8.

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

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

Second term: 6*(-3/8) = -18/8 = -9/4.

Third term: 9/4.

Total: 9/8 - 9/4 + 9/4 = 9/8.

Yes, that's correct.

So, that seems to check out.

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 opens downward, 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 if there's another way to approach this problem, maybe by considering partial derivatives.

Alternatively, since Archimedes chooses x first, and then Brahmagupta chooses y, we can think of this as a two-stage optimization.

But I think the method I used is correct.

Alternatively, maybe we can think of the problem as a game where we need to find the saddle point.

But since Archimedes moves first, it's not a symmetric game.

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

But since we already have the expression for the maximum z for each x, which is 8x² + 6x + 9/4, and we found its minimum at x = -3/8, that's sufficient.

Alternatively, let me check if there's a possibility that Brahmagupta might not choose the critical point if there are constraints on y? But the problem doesn't mention any constraints on x or y, so we can assume they can be any real numbers.

Therefore, the calculation is valid.

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

But let me check once more.

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

If we consider the game, Archimedes wants to choose x to minimize the maximum z over y.

Alternatively, maybe we can think of this as a function of x and y, and find the value where the minimum over x of the maximum over y is achieved.

But since we already did that, and verified with examples, I think it's safe.

Therefore, the value of x that Archimedes should choose is -3/8.

**Final Answer**
The value of $ x $ that Archimedes should choose is $\boxed{-\dfrac{3}{8}}$.

To determine the optimal value of $ x $ that Archimedes should choose, we analyze the function:

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

This is a two-player game where:

- **Archimedes** chooses $ x $ first, aiming to **minimize** $ z $.
- **Brahmagupta** chooses $ y $ second, aiming to **maximize** $ z $, given the chosen $ x $.

This is a classic **minimax problem**, where Archimedes wants to choose $ x $ such that the **maximum** value of $ z $ over all possible $ y $ is as small as possible.

---

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

We expand and simplify the function:

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

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 a fixed $ x $

For a quadratic $ ay^2 + by + c $, the vertex occurs at:

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

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

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

This is the value of $ y $ that **maximizes** $ z $ for a given $ x $.

---

### Step 3: Substitute $ y^* $ back into $ z(x, y) $ to find the maximum value of $ z $ for each $ x $

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

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

Compute each term:

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

Now combine all terms:

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

Convert all terms to have denominator 4:

- $ \frac{(8x + 3)^2}{4} $
- $ -\frac{(4x + 3)^2}{2} = -\frac{2(4x + 3)^2}{4} $
- $ \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 $
- $ (4x + 3)^2 = 16x^2 + 24x + 9 $
- $ 6(4x + 3) = 24x + 18 $

Substitute:

$$
z(x, y^*) = \frac{64x^2 + 48x + 9 - 2(16x^2 + 24x + 9) + 24x + 18}{4}
$$

Simplify numerator:

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

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**:

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

---

### 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 because this seems straightforward, but I want to make sure I didn't make a mistake in the earlier steps.

Let me recap:

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

For fixed x, found y* = -(4x + 3)/2.

Substituted back into z(x, y) and after simplifying, got z = 8x² + 6x + 9/4.

Then, minimized that quadratic, getting x = -3/8.

But let me verify the substitution step again because that's where errors can creep in.

So, let me recompute z at y*:

First term: (2x - y*)².

Earlier, I found 2x - y* = (8x + 3)/2.

So, squared is (8x + 3)^2 / 4.

Second term: -2y*².

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

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

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

So, adding all three terms:

(8x + 3)^2 / 4 - (4x + 3)^2 / 2 + 3*(4x + 3)/2.

Let me compute each term again:

First term: (8x + 3)^2 / 4 = (64x² + 48x + 9)/4.

Second term: -(4x + 3)^2 / 2 = -(16x² + 24x + 9)/2.

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

Now, converting all to quarters:

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.

Adding them together:

Numerator: 64x² + 48x + 9 - 32x² - 48x - 18 + 24x + 18.

Compute term by term:

64x² - 32x² = 32x².

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

9 - 18 + 18 = 9.

So numerator is 32x² + 24x + 9. Divided by 4: (32x² + 24x + 9)/4 = 8x² + 6x + 9/4.

Yes, that's correct.

So, the expression is indeed 8x² + 6x + 9/4.

Therefore, the minimum occurs at x = -b/(2a) = -6/(2*8) = -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 game theory aspect.

Alternatively, since Archimedes chooses x first, and Brahmagupta chooses y second, we can think of this as a two-player game where the value is z(x, y). The minimax value is the minimum over x of the maximum over y of z(x, y).

So, the value we found, 8x² + 6x + 9/4, is the maximum over y of z(x, y) for each x. Then, Archimedes minimizes this.

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

But since we already did that, and the result seems consistent, maybe that's okay.

But let me check with another approach. Let me think about the original function.

Alternatively, maybe we can consider the entire game as a function of x and y, and find the saddle point.

But since Archimedes chooses x first, and then Brahmagupta chooses y, the optimal strategy for Archimedes is to choose x such that the maximum over y of z(x, y) is minimized.

Alternatively, maybe we can think of this as a zero-sum game, and find the value of the game.

But perhaps my initial approach is correct.

But let me check with specific values. Let me pick x = -3/8 and see what y Brahmagupta would choose, and compute z.

So, x = -3/8. Then, y* = -(4x + 3)/2.

Compute 4x + 3: 4*(-3/8) + 3 = -12/8 + 3 = -3/2 + 3 = 3/2.

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

So, y = -3/4.

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

First, 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.

Now, let me check if this is indeed the minimum. Let me pick another x, say x = 0.

For x = 0, y* = -(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.

Which is 2.25, while 9/8 is 1.125, so indeed lower.

Another test: x = -3/8, z = 9/8.

What if x = -1/2? Let's compute.

x = -1/2.

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

Compute z(-1/2, -1/2):

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

= (-1 + 1/2)^2 - 2*(1/4) + 3/2

= (-1/2)^2 - 1/2 + 3/2

= 1/4 - 1/2 + 3/2 = 1/4 + 1 = 5/4 = 1.25.

Which is 1.25, which is higher than 9/8 = 1.125.

So, that seems consistent.

Another test: x = -3/8.

Alternatively, let me check the expression 8x² + 6x + 9/4 at x = -3/8.

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

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

Second term: 6*(-3/8) = -18/8 = -9/4.

Third term: 9/4.

Total: 9/8 - 9/4 + 9/4 = 9/8.

Yes, that's correct.

So, that seems to check out.

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 opens downward, 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 if there's another way to approach this problem, maybe by considering partial derivatives.

Alternatively, since Archimedes chooses x first, and then Brahmagupta chooses y, we can think of this as a two-stage optimization.

But I think the method I used is correct.

Alternatively, maybe we can think of the problem as a game where we need to find the saddle point.

But since Archimedes moves first, it's not a symmetric game.

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

But since we already have the expression for the maximum z for each x, which is 8x² + 6x + 9/4, and we found its minimum at x = -3/8, that's sufficient.

Alternatively, let me check if there's a possibility that Brahmagupta might not choose the critical point if there are constraints on y? But the problem doesn't mention any constraints on x or y, so we can assume they can be any real numbers.

Therefore, the calculation is valid.

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

But let me check once more.

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

If we consider the game, Archimedes wants to choose x to minimize the maximum z over y.

Alternatively, maybe we can think of this as a function of x and y, and find the value where the minimum over x of the maximum over y is achieved.

But since we already did that, and verified with examples, I think it's safe.

Therefore, the value of x that Archimedes should choose is -3/8.

**Final Answer**
The value of $ x $ that Archimedes should choose is $\boxed{-\dfrac{3}{8}}$.

To determine the optimal value of $ x $ that Archimedes should choose, we analyze the function:

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

This is a two-player game where:

- **Archimedes** chooses $ x $ first, aiming to **minimize** $ z $.
- **Brahmagupta** chooses $ y $ second, aiming to **maximize** $ z $, given the chosen $ x $.

This is a classic **minimax problem**, where Archimedes wants to choose $ x $ such that the **maximum** value of $ z $ over all possible $ y $ is as small as possible.

---

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

We expand and simplify the function:

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

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 a fixed $ x $

For a quadratic $ ay^2 + by + c $, the vertex occurs at:

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

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

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

This is the value of $ y $ that **maximizes** $ z $ for a given $ x $.

---

### Step 3: Substitute $ y^* $ back into $ z(x, y) $ to find the maximum value of $ z $ for each $ x $

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

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

Compute each term:

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

Now combine all terms:

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

Convert all terms to have denominator 4:

- $ \frac{(8x + 3)^2}{4} $
- $ -\frac{(4x + 3)^2}{2} = -\frac{2(4x + 3)^2}{4} $
- $ \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 $
- $ (4x + 3)^2 = 16x^2 + 24x + 9 $
- $ 6(4x + 3) = 24x + 18 $

Substitute:

$$
z(x, y^*) = \frac{64x^2 + 48x + 9 - 2(16x^2 + 24x + 9) + 24x + 18}{4}
$$

Simplify numerator:

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

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**:

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

---

### 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 $x = -\frac{6}{16} = \boxed{-\frac{3}{8}}.$