# Find length and width

• Oct 25th 2009, 06:47 PM
Nightasylum
Find length and width
of a rectangle that has the given perimeter and a maximum area.

Perimeter : 80 meters
Im still not sure how to set these up to prove with calculus.
2x+2y=80?
• Oct 25th 2009, 06:52 PM
Chris L T521
Quote:

Originally Posted by Nightasylum
of a rectangle that has the given perimeter and a maximum area.

Perimeter : 80 meters
Im still not sure how to set these up to prove with calculus.

Let a rectangle have length $l$ and width $w$.

Then by the first condition, we have $P=2l+2w=80$.

Let $A=lw$. It follows from above that $l=40-w$.

Substituting this into the area equation, we have $A=(40-w)w=40w-w^2$.

Since we want to maximize area, it follows that $\frac{\,dA}{\,dw}=40-2w$. Now, we find the critical point:

$40-2w=0\implies w=20$ (To verify its a max, we differentiate again to get $A^{\prime\prime}=-2<0$)

Since $w=20$, it follows that $l=40-(20)=20$.

Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.

Does this make sense?
• Oct 25th 2009, 06:57 PM
Nightasylum
Quote:

Originally Posted by Chris L T521
Let a rectangle have length $l$ and width $w$.

Then by the first condition, we have $P=2l+2w=80$.

Let $A=lw$. It follows from above that $l=40-w$.

Substituting this into the area equation, we have $A=(40-w)w=40w-w^2$.

Since we want to maximize area, it follows that $\frac{\,dA}{\,dw}=40-2w$. Now, we find the critical point:

$40-2w=0\implies w=20$ (To verify its a max, we differentiate again to get $A^{\prime\prime}=-2<0$)

Since $w=20$, it follows that $l=40-(20)=20$.

Therefore, the dimensions of the rectangle that gives us the maximum area actually form a square.

Does this make sense?

Yes perfect sense. Thanks alot