Question

Find the dimensions of a rectangle with perimeter 116 m whose area is as large as possible.

Answer 1

make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square make a square

Answer 2

2x+2y=116
where x is length and y is width

A=xy


solve for x or y from 2x+2y=116
doesn't matter which
go ahead and divide both sides by 2 before doing so
x+y=58
so we have y=58-x


now A=x(58-x)=58x-x^2
A'=58-2x
58-2x=0
2x=58
x=58/2=29

so y=29

Answer 3

calculus or algebra??

Answer 4

Calc

Answer 5

ok cool
do you understand?

Answer 6

Yes, myininaya. Thank you so much, very good explanation! Medal earned without question.

Answer 7

\[P=2l+2w\]
\[\frac{P}{2}=l+2\]
\[\frac{P}{2}-l=w\]
\[A=l\times w\]
\[A(l)=l(\frac{P}{2}-l)\]
\[A(l)=\frac{Pl}{2}-l^2\]

Answer 8

a parabola that faced down, vertex is
\[\frac{-b}{2a}=\frac{P}{4}\] and so you see you have a square

Answer 9

btw of course myininaya is correct. however it is entirely unnecessary to use calc to find the vertex of a parabola.