Absolute maximum for $f(x,y) = 4x^2 + 5y$

Question

Suppose you have a region $R$ that satisfies $4x^2+y^2\leq 4$ What is the absolute maximum of the function $f(x,y) = 4x^2 + 5y$ on $R$?

The correct answer is $10$

First, I calculated the critical points in $R$:

$$f_x = 8x$$ $$8x= 0$$ $$x=0$$

$$f_y = 5$$ $$5 \ne 0$$

Given the fact that there are no points where both $f_x$ and $f_y$ equal $0$, doesn't that mean there are no critical points in $R$, indicating no maximum value? I don't know if I have to check the boundary in this case, since the region is circular? I am not allowed to use Lagrange Multipliers.

Any insight into how to solve these types of problems would help.

Answer 1

The function $f(x)=x$ also has an absolute maximum on $[-1,1]$, even though there is no point at which $f'(x)=0$.

Remember, the zeroes of the partial derivatives only give you candidates for the critical points in the interior of the domain of $f$. On the border (where a maximum can also be reached sometimes), you can use Lagrange multipliers to calculate the max.

Alternatively, your particular function is fairly simple, so you can do without Lagrange. You have, on the border, $4x^2+y^2=4$, which means you can express $x^2$ as a funciton of $y$, and insert that into $f(x,y)$ to get a function of only $y$, and you can calculate the maximum of $f$ on the border quite easily.

Answer 2

You proved that there are no critical points in the interior of $R$. In order to see what happens at the boundary of $R$, you can use the method of Lagrange multipliers, which leads you to the system:$$\left\{\begin{array}{l}8x=8\lambda x\\5=8\lambda y\\4x^2+y^2=4.\end{array}\right.$$

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Otherwise, you can reduce this to a $1$-dimensional problem: find the maximum of$$(4\cos\theta)^2+5\times4\sin\theta,$$with $\theta\in[0,2\pi]$. It is perhaps better to keep in mind that $\cos^2\theta+\sin^2\theta=1$.

Answer 3

When the function is defined by a simple expression ad-hoc methods are recommended. You can find the maximum without using any theorems. $f(x,y) \leq 5y+4-y^{2}$ and $y$ is subject to the condition $-2\leq y \leq 2$. This gives $f(x,y)\leq 10$ and this value is attained when $x=0,y=2$.

Answer 4

We have: $4x^2 + 5y \le 4-y^2+5y =10 -y^2+5y-6=10 -(y-2)(y-3)\le 10$ since $y \le 2$.