To minimize the function $Q = 5x^2 + 3y^3$ subject to the constraint $x + y = 8$, we can use the method of Lagrange multipliers.

Step 1: Define the Lagrange function

We introduce a Lagrange multiplier $\lambda$ and define the Lagrange function as: $\mathcal{L}(x, y, \lambda) = 5x^2 + 3y^3 + \lambda (x + y - 8)$

Step 2: Take partial derivatives

Next, we take the partial derivatives of $\mathcal{L}(x, y, \lambda)$ with respect to $x$, $y$, and $\lambda$, and set them equal to zero.

1. $\frac{\partial \mathcal{L}}{\partial x} = 10x + \lambda = 0$ $10x + \lambda = 0 \quad \Rightarrow \quad \lambda = -10x$

2. $\frac{\partial \mathcal{L}}{\partial y} = 9y^2 + \lambda = 0$ $9y^2 + \lambda = 0 \quad \Rightarrow \quad \lambda = -9y^2$

3. $\frac{\partial \mathcal{L}}{\partial \lambda} = x + y - 8 = 0$ $x + y = 8$

Step 3: Set the Lagrange multipliers equal

Since both expressions for $\lambda$ must be equal, we set them equal to each other: $-10x = -9y^2 \quad \Rightarrow \quad 10x = 9y^2$ Thus, $x = \frac{9y^2}{10}$

Step 4: Substitute into the constraint

Now, substitute $x = \frac{9y^2}{10}$ into the constraint $x + y = 8$: $\frac{9y^2}{10} + y = 8$ Multiply the entire equation by 10 to eliminate the denominator: $9y^2 + 10y = 80$ This simplifies to the quadratic equation: $9y^2 + 10y - 80 = 0$

Step 5: Solve the quadratic equation

We solve the quadratic equation $9y^2 + 10y - 80 = 0$ using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 9$, $b = 10$, and $c = -80$.

Substituting these values: $y = \frac{-10 \pm \sqrt{10^2 - 4(9)(-80)}}{2(9)} = \frac{-10 \pm \sqrt{100 + 2880}}{18} = \frac{-10 \pm \sqrt{2980}}{18}$

Thus, $y = \frac{-10 \pm \sqrt{2980}}{18}$

Step 6: Calculate $x$

Once you get the values of $y$, you can substitute them back into $x = \frac{9y^2}{10}$ to find the corresponding values of $x$.

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