Can anyone tell me if this is correct?

Question

Suppose that the temperature of a metal plate is given by $T(x; y) = x^2 +2x+y^2$, for points $(x, y)$ on the elliptical plate defined by $x^2 + 4y^2 <= 24$. Find the maximum and minimum temperatures on the plate.

This is what i have done so far.

Finding critical point: $T(x)=2x+2$, $T(y)=2y$. Equating to $0$, $x=-1$, $y=0$. Critical point is $(-1,0)$ and is a minimum.

On the boundary, $ x^2 + 4y^2 = 24$

$g(x,y)=x^2 + 4y^2$

$g(x)=2x, g(y)=8y$

$2x+2=A2x$---------(1)

$2y =A8y$---------(2)

$x^2+4y^2=24$------(3)

When solving from equation 1 and 3 im getting $ x=-1,y=|(23/4)^{0.5}|,$ and $x=|24^{0.5}|, y=0$ and when from eqn 2 and 3 im getting $x=|24^{0.5}|,y=0, A= 0.25 , x=-4/3, y=|(50/9)^{0.5}|$.

Is this correct? am getting different values when using equation $1$ and $2$.

Answer 1

hint:Equation $(2)$ gives $A = 1/4$, or $y = 0$. If $A = 1/4 \to 2x+2 = \dfrac{x}{2}\to x = -4/3$.

If $A = 2, (2) \to y = 0, (1) \to x = 1 \to x^2+4y^2 = 1^2 + 4\cdot 0^2 = 1 \neq 24 \to A \neq 2$.

Answer 2

For minimum, point $(-1,0)$ should be correct.

Although I didn't understand your approach for finding maximum, maximum temperature will occur at the boundaries.

$x^2 + 4y^2 = 24$

$y^2 = \frac{24-x^2}{4}$

Substituting this value into the equation for temperature yields $T(x)= \frac{3x^2+8x+24}{4}$

As $x$ ranges from $-\sqrt{24} \space to \space \sqrt{24}$.

$x=\sqrt{24} $ gives $33.798$ as the maximum.