# Thread: Find length and width

1. ## 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?

2. 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 $\displaystyle l$ and width $\displaystyle w$.

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

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

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

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

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

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

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

Does this make sense?

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

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

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

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

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

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

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

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

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