How to find absolute maximum of a bivariate quadratic function subject to constraints?

Question

The question I am trying to solve is

$$\begin{array}{ll} \text{extremize} & f(x,y) := xy − 6y − 36x + 216\\ \text{subject to} & y \geq x^2\\ & y \leq 37\end{array}$$

Now I know how to solve these types of questions when the constraints are inequalities on $x$ and $y$, such as $0 \leq x \leq 5$ and $2 \leq y \leq 6$. First, we find the interior critical points and then we find the boundary critical points. Similarly, first, I calculated interior CP and I only got $f(6,36)=0$. Then, I took $x= -\sqrt{37}$ into $f(x,y)$ and got a function of $y$. Then I put in $y=37$ to find the minimum value. It turned out to be correct. The minimum value is $f(-\sqrt{37}, 37) = -12.08$.

But when I take $x=\sqrt{37}$ and $y=37$, value of $f(x,y)$ is $0.08$ and it turns out to be incorrect. I do not know what I am doing wrong and what other values of $x$ or $y$ to take to get another critical point of $f(x,y)$.

Answer 1

Guide:

• As you mentioned, the optimal solution can occur in the interior or exterior or on the boundary.

• You have computed the critical point for the interior.

• Now we can focus on the boundary, an optimal solution need not be at the vertices.

• For the $y=37$ boundary, you can first let $y=37$ and $-\sqrt{37} \le x \le \sqrt{37}$ and see where are the optimal points.

• For $y=x^2$, substitute $y=x^2$ inside the original objective function and solve for $f(x)=x^3-6x^2-36x+216$ where $-\sqrt{37} \le x \le \sqrt{37}$.
• Compare all possible optimal value to get the maximum and minimum.

Answer 2

At first you get $$\frac{\partial f(x,y)}{\partial x}=y-36$$ and $$\frac{\partial f(x,y)}{\partial y}=x-6$$

Then you have for $$-\sqrt{37}\le x\le \sqrt{37}$$ to plug into your function $$y=37$$ and for $$-\sqrt{37}\le x\le \sqrt{37}$$ plug into your function $y=x^2$

Answer 3

Since $f$ has no critical points in the interior of the region that you described, you must see what happens at th boundary. Now, consider the points $(x,y)$ with $y=x^2$ and $|x|\leqslant\sqrt{37}$. In other words, consider $g(x)=f(x,x^2)=x^3-6x^2-36x+216$. Then$$g'(x)=3x^2-12x-36=3(x+2)(x-6).$$So, $g$ has a local maximum at $-2$, which is in fact an absolute maximum, since $x\in\left[-\sqrt{37},\sqrt{37}\right]$. Therefore, the maximum of $f$ is $g(-2)=256$.

Answer 4

Apply the Lagrange multipliers technique introducing slack variables to operate with equality restrictions

Calling

$$ f(x,y) = x y -6 y -36 x+216\\ g_1(x,y,\epsilon) = y-x^2-\epsilon^2\\ g_2(x,y,\epsilon) = y-37-\epsilon^2 $$

and then

$$ L(x,y,\lambda_i,\epsilon_i)=f(x,y) + \sum_{i=1}^2 \lambda_ig_i(x,y,\epsilon_i) $$

so the stationarity condition reads

$$ \left\{ \begin{array}{rcl} y-2\lambda_1-36 & = & 0\\ x+\lambda_1+\lambda_2-6 & = & 0\\ y-x^2-\epsilon_1^2 & = & 0\\ y+\epsilon_2^2-37 & = & 0\\ \lambda_1\epsilon_1 & = & 0\\ \lambda_2\epsilon_2 & = & 0 \end{array} \right. $$

Now solving this system we determine the interior and boundary stationary points.

In this case the solution gives

$x =-\sqrt{37}, y = 37, \lambda_1 = -(1/(2 \sqrt{37})), \lambda_2 = 3/74 (148 + 25 \sqrt{37}), \epsilon_1 = 0, \epsilon_2 = 0$ giving $f(x,y) = -(6+\sqrt{37})$ which is a boundary point because $\epsilon_1 = \epsilon_2 = 0$