Minimum and Maximums involving Partial Derivatives

Question

Hi all. I was wondering if someone could help me learn to approach this problem.

the partial derivatives I have are:

$f_x=4x-8$

$f_y=2y$

$f_{xy} = 0.$

How does the constraint affect the problem? Are my partial derivatives correct?

Answer 1

Using the theorem:

Let $f$ be a function with two variables with continuous second order partial derivatives $f_{xx}, f_{yy}, f_{xy}$ at a critical point $(a,b)$ $$D=f_{xx}(a,b)f_{yy}(a,b)-f_{xy}^2(a,b)$$

• If $D > 0$ and $f_{xx}(a,b) > 0$, then $f$ has a relative minimum at $(a,b)$.
• If $D > 0$ and $f_{xx}(a,b) < 0$, then $f$ has a relative maximum at $(a,b)$.

$$f_x=4x-8 \Rightarrow f_{xx}=4, \\ f_y=2y \Rightarrow f_{yy}=2, \\ f_{xy}=0$$

The critical points satisfy the equations $f_x(x,y)=0$ and $f_y(x,y)=0$ simultaneously.

Therefore, $f_x=4x-8 =0 \Rightarrow x=2$ and $f_y=2y=0 \Rightarrow y=0$

So the critical point $(a,b)$ is $(2,0)$.

This point satisfies the contraint: $2^2+0^2=4 \leq 9$.

$$D=f_{xx}(2,0)f_{yy}(2,0)-f_{xy}^2(2,0)=4 \cdot 2=8>0$$

Since $D>0$ and $f_{xx}(2,0)>0$ the function $f$ has a local minimum at $(2,0)$, which is equal to $f(2,0)=-8$.

EDIT:

The problem $$\max 2x^2-8x+y^2 \\ x^2+y^2 -9\leq 0$$ is a nonlinear programming problem.

So we can use the following:

$$\max f(X) \\ g_i(X) \leq 0, \ \ \ i=1, \dots, p $$

($X$ is a vector: $X=(X_1, X_2, \dots, X_n)$)

If there are $X^* \in \mathbb{R}^n, \mu \in \mathbb{R}^p$ such that $$ \mu_k \geq 0 , k =1,2, \dots, p \\ \frac{\partial{f(X^*)}}{\partial{X_j}}-\sum_{k=1}^p \mu_k \frac{\partial{g_k(X^*)}}{\partial{X_j}}=0, j=1,2, \dots , n \\ \mu_k g_k(X^*) =0, k=1,2, \dots, p \\ g_k(X^*) \leq 0, k=1,2, \dots, p$$ then $X^*$ is the solution of the problem, that means that it is the maximum of $f$ subject to the constraint $g(X) \leq 0$.

Is this case the vector $X$ is $(x,y)$ and $g(x,y)=x^2+y^2-9$

$n=2$, $p=1$

So, we have the following:

$$ \mu_1 \geq 0 \ \ \ (1) \\ 4x-8- \mu_12x=0 \ \ \ (2) \\ 2y-\mu_1 2y=0 \ \ \ (3) \\ \mu_1 (x^2+y^2-9) =0 \ \ \ (4) \\ x^2+y^2-9 \leq 0 \ \ \ (5) $$

$(3) \Rightarrow 2y(1- \mu_1)=0 \Rightarrow y=0 \text{ or } \mu_1=1$

• $\mu_1=1:$

$(2) \Rightarrow 4x-8-2x=0 \Rightarrow 2x=8 \Rightarrow x=4$

$(4) \Rightarrow 16+y^2-9 =0 \Rightarrow y^2=-5$ That cannot be true.

Therefore, it should be $y=0$.

• $y=0:$

Then relations are as followed:

$$ \mu_1 \geq 0 \ \ \ (1) \\ 4x-8- \mu_12x=0 \ \ \ (2) \\ \mu_1 (x^2-9) =0 \ \ \ (4) \\ x^2-9 \leq 0 \ \ \ (5) $$

$(4) \Rightarrow \mu_1=0 \text{ or } x= \pm 3$

If $\mu_1=0:$

$(2) \Rightarrow 4x-8=0 \Rightarrow x=2$

$(5) \Rightarrow 4-9 =-5 \leq 0\checkmark$

So, since all the $5$ relations are satisfied $(2,0)$ is a point of possible maximum. The value of $f$ at this point is $f(2,0)=-8$.

If $x=3:$

$(2) \Rightarrow 12-8- \mu_1 6=0 \Rightarrow 6 \mu_1=4 \Rightarrow \mu_1=\frac{2}{3}$

So, since all the $5$ relations are satisfied $(3,0)$ is also a point of possible maximum. The value of $f$ at this point is $f(3,0)=-6$.

Since $-6>-8$ the previous one $f(2,0)$ is not the maximum.

If $x=-3:$

$(2) \Rightarrow -12-8+ \mu_1 6= \Rightarrow 6 \mu_1=20 \Rightarrow \mu_1=\frac{10}{3}$

So, since all the $5$ relations are satisfied $(-3,0)$ is also a point of possible maximum. The value of $f$ at this point is $f(-3,0)=42$.

Since $42>-6$ the previous one $f(3,0)$ is not the maximum.

So, we conclude that the function $f$ achieves its maximum at the pooint $(-3, 0)$, which is equal to $f(-3,0)=42$.