Okay, so Archimedes and Brahmagupta are playing this game with the function z(x, y) = (2x - y)^2 - 2y² - 3y. Archimedes goes first, choosing x, and then Brahmagupta chooses y. Archimedes wants to minimize z, while Brahmagupta wants to maximize it. We need to find the value of x that Archimedes should choose, assuming Brahmagupta plays optimally. Hmm, this sounds like a minimax problem. Let me think through this step by step.

First, I need to understand the structure of the function z(x, y). It's a quadratic function in both x and y. Let me rewrite it to see if I can simplify it or complete the square. The function is:

z = (2x - y)² - 2y² - 3y.

Let me expand the first term:

(2x - y)² = 4x² - 4xy + y².

So substituting back into z:

z = 4x² - 4xy + y² - 2y² - 3y.

Combine like terms:

4x² - 4xy + (y² - 2y²) - 3y = 4x² - 4xy - y² - 3y.

So z = 4x² - 4xy - y² - 3y.

Hmm, okay. Now, since Archimedes chooses x first, and then Brahmagupta chooses y, given x. So Archimedes is trying to choose an x such that, whatever Brahmagupta does in response, z is minimized. But Brahmagupta will choose y to maximize z, given that x was chosen by Archimedes. So it's a sequential game with Archimedes choosing x to minimize the maximum z that Brahmagupta can achieve.

This is a minimax problem. So the value for Archimedes is the minimum over x of the maximum over y of z(x, y). So we need to find x that minimizes the maximum z(Brahmagupta's choice of y given x).

So first, for a given x, Brahmagupta will choose y to maximize z. Then, Archimedes chooses x to minimize that maximum z.

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

Let me start with step 1. For a given x, find y to maximize z.

Given z = 4x² - 4xy - y² - 3y.

This is a quadratic in y. Let's write it as a function of y:

z(y) = -y² - (4x + 3)y + 4x².

It's a quadratic function in y, and since the coefficient of y² is -1 (negative), the parabola opens downward. Therefore, the maximum occurs at the vertex.

The vertex of a quadratic ay² + by + c is at y = -b/(2a). Here, a = -1, b = -(4x + 3). So the maximum y is at y = -b/(2a) = -[-(4x + 3)]/(2*(-1)) = (4x + 3)/(-2) = -(4x + 3)/2.

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

Wait, let me verify that. The standard formula for the vertex is y = -B/(2A) where the quadratic is Ay² + By + C. In our case, z(y) = -1*y² + (-4x -3)*y + 4x². So A = -1, B = -4x -3.

So the vertex is at y = -B/(2A) = -(-4x -3)/(2*(-1)) = (4x + 3)/(-2) = -(4x + 3)/2. Yes, that's correct. So Brahmagupta will choose y = -(4x + 3)/2 to maximize z.

Now, substitute this y back into z to find the maximum z for each x.

So z_max(x) = z(x, y) where y = -(4x + 3)/2.

Let me compute that:

z = 4x² - 4x*y - y² - 3y.

Substituting y = -(4x + 3)/2:

First, compute each term:

4x² remains as is.

-4x*y = -4x*(-(4x + 3)/2) = 4x*(4x + 3)/2 = 2x*(4x + 3) = 8x² + 6x.

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

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

So now, sum all these up:

4x² + 8x² + 6x - (16x² + 24x + 9)/4 + (12x + 9)/2.

Combine like terms:

First, combine the x² terms:

4x² + 8x² = 12x².

Then the linear terms:

6x.

Then the constants and other terms:

- (16x² + 24x + 9)/4 + (12x + 9)/2.

Let me handle the fractions step by step.

First, expand each term:

-16x²/4 -24x/4 -9/4 + 12x/2 + 9/2.

Simplify each:

-4x² -6x -9/4 + 6x + 9/2.

Now combine like terms:

-4x².

-6x + 6x = 0.

-9/4 + 9/2 = -9/4 + 18/4 = 9/4.

So putting it all together:

12x² (from the first two terms) + 0 (from linear terms) + 9/4.

Therefore, z_max(x) = 12x² + 9/4.

Wait, that seems too simple. Let me check the calculations again.

Starting from substituting y = -(4x + 3)/2 into z:

z = 4x² -4x*y - y² -3y.

Compute each term:

-4x*y = -4x*(-(4x + 3)/2) = 4x*(4x + 3)/2 = 2x*(4x + 3) = 8x² + 6x. Correct.

-y² = -[( -(4x + 3)/2 )²] = -[(16x² + 24x + 9)/4] = -4x² -6x -9/4. Wait, I think I made a mistake here. Let me recalculate that.

Wait, y = -(4x + 3)/2. So y² = [ (4x + 3)/2 ]² = (16x² + 24x + 9)/4. Therefore, -y² = - (16x² + 24x + 9)/4. So that term is -4x² -6x -9/4. Wait, no. Wait, (16x² +24x +9)/4 is 4x² +6x +9/4. So -y² is -4x² -6x -9/4. Correct.

Similarly, -3y = -3*(-(4x +3)/2) = (12x +9)/2. Correct.

So putting all together:

4x² (from 4x²) + (8x² +6x) (from -4x*y) + (-4x² -6x -9/4) (from -y²) + (12x +9)/2 (from -3y).

So let's combine term by term:

4x² +8x² -4x² = 8x².

6x -6x +12x = 12x.

-9/4 +9/2. Which is -9/4 +18/4 =9/4.

So total z_max(x) =8x² +12x +9/4.

Wait, but when I did the initial calculation, I thought it was 12x² +9/4. But here, it's 8x² +12x +9/4. Which is different. So I must have made a mistake earlier.

Wait, let me re-express step by step:

Original expression after substitution:

z =4x² +8x² +6x -4x² -6x -9/4 +6x +9/2.

Wait, hold on. Let me re-express all terms:

4x² (from 4x²)

+8x² +6x (from -4x*y)

-4x² -6x -9/4 (from -y²)

+6x +9/2 (from -3y)

So now, combining like terms:

x² terms: 4x² +8x² -4x² =8x².

x terms:6x -6x +6x=6x.

Constants: -9/4 +9/2= -9/4 +18/4=9/4.

Therefore, z_max(x)=8x² +6x +9/4.

Wait, so that's 8x² +6x +9/4. So that's a quadratic in x. So now, Archimedes wants to choose x to minimize this expression. So we need to find the x that minimizes 8x² +6x +9/4.

Since this is a quadratic in x with a positive coefficient for x², 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, the value of x that Archimedes should choose is -3/8.

But wait, let me verify this again. Let's confirm the steps.

1. Expressed z in terms of x and y: z =4x² -4xy -y² -3y.

2. For a given x, z is a quadratic in y. Completed the square or found vertex at y = -(4x +3)/2.

3. Substituted this y into z, leading to z_max(x) =8x² +6x +9/4.

4. Then, minimized this quadratic in x, which is 8x² +6x +9/4. The minimum is at x = -6/(2*8) = -3/8.

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

Wait, but let me check the substitution again to make sure I didn't make a mistake.

Original z after substitution:

z =4x² -4x*y -y² -3y.

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

Compute each term:

4x².

-4x*y = -4x*(-(4x +3)/2) = 2x*(4x +3) =8x² +6x.

-y² = -[( -(4x +3)/2 )²] = -[(16x² +24x +9)/4] = -4x² -6x -9/4.

-3y = -3*(-(4x +3)/2) = (12x +9)/2 =6x +9/2.

Adding all these:

4x² +8x² +6x -4x² -6x -9/4 +6x +9/2.

Yes, 4x² +8x² -4x² =8x².

6x -6x +6x=6x.

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

So total z_max(x)=8x² +6x +9/4. Correct.

Then, finding the minimum of 8x² +6x +9/4. The derivative is 16x +6. Setting to zero: 16x +6=0 =>x= -6/16= -3/8. Correct.

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

Wait, but let me check if there's another way to approach this problem, perhaps by completing the square for the original function.

Original function: z=4x² -4xy -y² -3y.

Alternatively, maybe we can complete the square in terms of y, which we did, then express z as a quadratic in x, but since x is chosen first, maybe there's another perspective.

Alternatively, treating z as a function of x and y, and considering the minimax.

But in this case, since for each x, we can find the optimal y, which is the vertex of the quadratic in y, and substitute back into z, resulting in a quadratic in x, which is then minimized. So the steps seem correct.

But let me verify once again.

Suppose x = -3/8. Then 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:

z = (2x - y)^2 -2y² -3y.

Compute 2x - y: 2*(-3/8) - (-3/4) = -6/8 +3/4 = -3/4 +3/4 =0. So (2x - y)^2 =0.

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

So z=0 +9/8=9/8. So z=9/8.

Wait, but according to our previous calculation, z_max(x) when x=-3/8 is 8*(-3/8)^2 +6*(-3/8) +9/4.

Compute that:

8*(9/64) +6*(-3/8) +9/4.

8*(9/64)=72/64=9/8.

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

9/4.

So total: 9/8 -9/4 +9/4=9/8. Which matches the direct computation. So that's correct.

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

But let me check if there's any alternative approach. For example, could we write the original function in terms of x and y, and then find the saddle point?

Alternatively, perhaps represent the function in a matrix form and analyze it, but that might be more complicated.

Alternatively, think of the game as a simultaneous move game, but since Archimedes chooses x first, then Brahmagupta responds, it's a sequential game with perfect information. So the solution is indeed as above.

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

**Final Answer**
The value of \( x \) that Archimedes should choose is \boxed{-\dfrac{3}{8}}.
Therefore, Archimedes should choose the value of \( x \) to be $\boxed{-\frac{3}{8}}$.
</think>