Critical points of a function with constraints.

Question

Could somebody help me with this exercise please?

Let $f$ be a function of two variables $y$ and $x$: $$f(x,y)=e^y(y^2-2x^2)$$

(1) Determine the maxima and minima of $f$ under the constraint $$g(x,y)=2x^2+y^2=6$$

(2) Determine the maxima and minima of $f$ on the set $$M=\{(x,y)\in\mathbb{R}^2|2x^2+y^2\leq 6\}$$

I would really appreciate your help.

Answer 1

1)

as you mentioned in the comments for $L_x=0$ we get $x=0$ or $\lambda=e^y$

for $x=0$:

$L_\lambda=2\cdot0^2+y^2-6=0 \rightarrow y=\pm \sqrt6$

so we have the points $(0,\sqrt6)$ and $(0,-\sqrt6)$

for $\lambda=e^y$:

$L_y=e^y(y^2-2x^2+2y+2y)=0$

so

$x^2=\frac{y^2+4y}{2}$

so we input $x^2$ in $L_\lambda$ we get:

$y^2+4y+y^2-6 \rightarrow y=-3$ and $y=1$

so we get the point $(\sqrt{\frac{7}{2}},1)$(no valid point for $y=-3$)

now we input all the points in $f(x,y)$:

we get the values: $6e^{\sqrt6},6e^{-\sqrt6}$ and $-\frac{5}{2}e$

so the maximum value is $6e^{\sqrt6}$ and minimum is $-\frac{5}{2}e$.

Answer 2

(1) We can simplify the constraint to $2x^2 = 6 - y^2$, and thus re-write the original function $f$ as $f(x, y) = e^y(2y^2 - 6)$. Now, as long as the constraint continues to hold true, we in effect have a single-variable function $f(y) = e^y(2y^2 - 6)$. To find the maxima and minima, use the multiplication rule to derive the function and set it equal to 0:

$$\frac{d}{dy} f(y) = e^y(4y) + (2y^2 - 6)e^y = e^y(2y^2 + 4y - 6) = 0$$

Since $e^y \neq 0$ for $\{y : R\}$, $2y^2 + 4y - 6 = 0$. Applying the quadratic formula, we get $\frac{-4 \pm \sqrt{16+4(2)6}}{4} = \frac{-4 \pm 8}{4} = 1, -3$. Graphing $\frac{d}{dy} f(y)$ we see that between $1$ and $-3$ the graph is negative, so $f(y)$ is decreasing in that range. Therefore, there is a relative minimum at $y=1$ and a relative maximum at $y=-3$.