Question

Show that among all rectangles with a given perimeter, the square has the largest area?

Answer 1

let L and W be the length and with of a rectangle.

we have 2W+2L=p (p some fixed number)

area \[A=L\cdot W\]

solve for W in fitst equation

\[W=\frac{p-2L}{2}\]

then \[A=L\frac{p-2L}{2}\]

\[A=\frac{1}{2}(Lp-2L^2)\]

\[\frac{dA}{dL}=\frac{1}{2}(p-4L)\]

set equal to zero ..solve


\[\frac{1}{2}(p-4L)=0\Rightarrow L=\frac{p}{4}\]

then

\[W=\frac{p-2L}{2}=\frac{p-2\frac{p}{4}}{2}=\frac{p-p/2}{2}=\frac{p/2}{2}=p/4\]

thus W=L and our rectangle is really a square.

Answer 2

one should also check that this is truly a max (could use 1st or 2nd derivative test...do this for all the problems you needed to do)

Answer 3

sorry, didn't catch that?

Answer 4

we need to determine if we found a maximum or minimum. this process will find both. the 1st or 2nd D-test will tell us if we found a max or min

Answer 5

ohh okay okay, got it! ;) thank you sooo much, you're a lifesaver! ;)

Answer 6

using the 2nd D-test

\[\frac{d^2A}{dL^2}=\frac{1}{2}(-4)=-2\]
this is true for all values of L even our critical number p/4

since the 2nd D is negative at our critical value we have found a max by the 2nd D-test

Answer 7

np

Answer 8

penpal this looks good.
a square always maximizes area given a perimeter

Answer 9

notice how its kinda the same process

Answer 10

yeah, i noticed, thanks for looking into it. you rock!