Extreme value problem

Protonics · New Member Offline Joined: 20 Sep 2009 Posts: 12

A mirror producer wants to start the production of a new series. The new mirrors has an area of 1 m^2 and are shaped like a rectangle with a semi circle in the left and right ends. There is going to be a frame around the entire mirror, and the crooked part of the frame is twice as expensive pr. square centimeter compared to the straight part. How big must the radius of the semi circles be in order to make the frame as cheap as possible to produce?

Thanks!

**Posted:** Mon Nov 02, 2009 5:27 am

Kent Merryfield

**Posted:** Mon Nov 02, 2009 12:07 pm

Protonics · New Member Offline Joined: 20 Sep 2009 Posts: 12

Yes its perimeter

**Posted:** Tue Nov 03, 2009 1:01 am

Protonics · New Member Offline Joined: 20 Sep 2009 Posts: 12

Thanks for your attention, but i managed to solve the problem myself.

**Posted:** Tue Nov 03, 2009 3:58 am

Kent Merryfield

The setup: let $x$ be the width and $y$ the height, so that the semicircles have radius $\frac y2.$

The area is $A=xy+\frac{\pi}4y^2.$

The cost is proportional to $C=2x+2\pi y.$

The problem: minimize $C$ subject to the constraint $A=1.$

By Lagrange multipliers:

$2=\lambda y$

$2\pi = \lambda\left(x+\frac{\pi}2y\right).$

Put these together to get

$2\pi=\frac{2}{y}\left(x+\frac{\pi}2y\right)$

$\pi y-\frac{\pi}2y=x$

$x=\frac{\pi}2y.$

Put that in the area constraint: $\frac{\pi}2y^2+\frac{\pi}4y^2=1.$

$y=\sqrt{\frac4{3\pi}}$ and $x=\sqrt{\frac{\pi}3}.$

This gives a cost of $C=2\sqrt{3\pi}.$

We still have to check that that's a minimum. Check the extremes. As $y\to 0,$ we get $x\to\infty$ and $C\to\infty.$ At the other end, we can set $x=0,$ getting $y=\sqrt{\frac4{\pi}}.$ That gives a cost of $C=4\sqrt{\pi}.$ Since $4>2\sqrt{3},$ the minimum value of $C$ does not occur in the boundary case, so our interior critical point is the minimum.

**Posted:** Tue Nov 03, 2009 8:23 am