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A circular hot plate given by the relationship $x^2 + y^2 \leq 4$ is heated according to the spatial temperature function $T(x,y) = 10 - x^2 + 2x - 4y^2$. Find the hottest temperatures on the plate and the points at which they occur.

I get $D=0$ and thats suppose to be F.M.L is that normal? Usually you are suppose to get $D<0$ or $D>0$ ....$D$ is the 2nd derivative test btw.

I got $Fx=-2x+2-2x\lambda$


$F \lambda= x^2+ y^2=4$

I am using Lagrange Multiplier...I hope I am not screwing up on the system of equations...ive done this over and over 4 times with different isolations...

I get $\lambda= -4, \quad x=-1/3, \quad y=1.97$

Kinda confused about where I am screwing up..

I am also confused with this question, they are all optimization...

[a]: mixedmath removed these links, but keeps this faux-comment for posterity

A rectangular topless box of volume $12$ m$^3$ is constructed such that the material for the back and bottom costs 2 times as much as the other three sides. What dimensions of the box will minimize the total cost?

With this Last question I am able to get $L*W*H=12$

Also I am able to get that if $Z=$ material for 1 side and $Y=$material for bottom...

then I have $5z+2y=$ Cost

Now I am lost because to use LaGrange multiplier I need a bound..

I have made the volume function my $G(W,h,L)$ and $k=12$ and let the cost function be my function to minimize and solve for

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I've never used "FML" in that way... also, you might look into using LaTex to make your equations look better. – The Chaz 2.0 Mar 22 '12 at 23:32
Go to meta. Type "latex" in the search bar. Also, you might visit – The Chaz 2.0 Mar 22 '12 at 23:39
up vote 1 down vote accepted

On the interior, we would look for the critical points of $T(x,y)$: $$ 0=\frac{\partial}{\partial x}T(x,y)=-2x+2\tag{1} $$ and $$ 0=\frac{\partial}{\partial y}T(x,y)=-8y\tag{2} $$ Thus, the only interior extremum would be $(1,0)$, where $T(x,y)=11$.

However, on the boundary, we get $$ \begin{align} T(2\cos(\theta),2\sin(\theta)) &=10-4\cos^2(\theta)+4\cos(\theta)-16\sin^2(\theta)\\ &=12\cos^2(\theta)+4\cos(\theta)-6 \end{align} $$ and $12u^2+4u-6$ reaches a minimum at $u=\cos(\theta)=-\frac16$, where $T(x,y)=-\frac{19}{3}$.

At the extremes of $\cos(\theta)$, $T(2,0)=10$ and $T(-2,0)=2$.

Thus, the extremes of $T$ are $11$ at $(1,0)$ and $-\frac{19}{3}$ at $(-\frac13,\frac{\sqrt{35}}{3})$ and $(-\frac13,-\frac{\sqrt{35}}{3})$.

Using Lagrange Multipliers

In the interior, there is no constraint, so we still get the interior extrema in the usual way; that is, $(1)$ and $(2)$.

However, on the boundary, we have the constraint $$ x^2+y^2=4\tag{3} $$ Thus, under the constraint $(3)$, we want to maximize and minimize $$ 10-x^2+2x-4y^2\tag{4} $$ Thus, we form the function $$ F(x,y,\lambda)=10-x^2+2x-4y^2+\lambda(x^2+y^2-4)\tag{5} $$ and find the points at which $$ 0=\frac{\partial}{\partial x}F(x,y,\lambda)=-2x+2+2\lambda x\tag{6} $$ and $$ 0=\frac{\partial}{\partial y}F(x,y,\lambda)=-8y+2\lambda y\tag{7} $$ We get $$ x=\frac{1}{1-\lambda}\tag{8} $$ and $$ y=0\textrm{ or }\lambda=4\tag{9} $$ Now $(8)$ and $(9)$ are still under the constraint $(3)$. So if $y=0$, then $x=\pm2$. If $\lambda=4$, then $x=-\frac13$ and $y=\pm\frac{\sqrt{35}}{3}$. Therefore, on the boundary, the extrema are contained in the points $$ \left\{(2,0),(-2,0),\left(-\frac13,\frac{\sqrt{35}}{3}\right),\left(-\frac13,-\frac{\sqrt{35}}{3}\right)\right\}\tag{10} $$ Checking $T(x,y)$ at these points, we get $\{10,2,-\frac{19}{3},-\frac{19}{3}\}$. Thus, the extrema on the boundary are $10$ and $-\frac{19}{3}$. Combined with the sole interior extremum of $11$, we get $11$ and $-\frac{19}{3}$ as the maximum and minimum of $T(x,y)$.

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can you please explain how you got that? I was following this video and got D=0 which is wrong: I know that in the first step you differentiated that function with respect to x then y for the 2nd equation but how do you know though which answers to take that are within boundary? like i dont get how to got all sorts of cos and sins in there – Raynos Mar 23 '12 at 0:27
using the method in vido i was able to get all the answers you have except the (1,0)...I am not sure why that wasnt a solution when i had the system of equations – Raynos Mar 23 '12 at 0:33
also to classif the extreme points u have to do 2nd derivative test for which i get lamda=-4 and for either Fxx or Fyy or Fxy i dont have any x or y in those only Lamdas and constants... – Raynos Mar 23 '12 at 0:40
ooo i see your suppose to check both max min at the function as well as on boundary...and that for boundary u use lagrange multiplier – Raynos Mar 23 '12 at 0:43
@Raynos: I will add to my answer regarding the use of Lagrange Multipliers. However, it looks as if you have found the difference between the constrained boundary and the unconstrained interior :-) – robjohn Mar 23 '12 at 0:47

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