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Mathematics 7 Online
OpenStudy (anonymous):

find the dimensions of a rectangle with an area of 108 square feet that has the minimum perimeter?

OpenStudy (anonymous):

Do you know the formula for area and perimeter?

OpenStudy (anonymous):

a=lw and perimeter 2l+2w ?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

We want to minimize perimeter, so we can use the area formula to find \(w\) in terms of \(l\).

OpenStudy (anonymous):

Remember that they told us \(a=108\).

OpenStudy (anonymous):

okay that's where I get lost idk whats after that

OpenStudy (anonymous):

Okay so \[ 108 = lw\\ p = 2l+2w \]

OpenStudy (anonymous):

Do you see how we can get rid of \(l\) by using substitution?

OpenStudy (anonymous):

not quite how would It be done

OpenStudy (anonymous):

\[ l = 108/w \]

OpenStudy (anonymous):

that's all

OpenStudy (anonymous):

Now substitute: \[ p=2(108/w)+2w \]

OpenStudy (anonymous):

In order to minimize \(p\) you need to find \(dp/dw\). It will tell you which width \(p\) has max/min at.

OpenStudy (anonymous):

You need to find \(dp/dw = 0\)

OpenStudy (anonymous):

wio your a genius how do I find dp/dw or is it 0 then where do I plug it in

OpenStudy (anonymous):

First you differentiate \(p\) in terms of \(w\).

OpenStudy (anonymous):

Can you differentiate: \[ p=2(108/w)+2w \]?

OpenStudy (anonymous):

p-2w=2(108/w) ?

OpenStudy (anonymous):

Find the derivative.

OpenStudy (anonymous):

Do you know how to differentiate?

OpenStudy (anonymous):

\[ p = 216w^{-1}+2w \]

OpenStudy (anonymous):

Just use power rule.

OpenStudy (anonymous):

then would it ;be p= 216/w2+2w =0?

OpenStudy (anonymous):

You don't know how to differentiate?

OpenStudy (anonymous):

not really

OpenStudy (anonymous):

\[ \frac{d(216w^{-1})}{dw} = -216w^{-2} \]

OpenStudy (anonymous):

\[ \frac{d(2w)}{dw} = 2 \]

OpenStudy (anonymous):

This gives us: \[ \frac{dp}{dw} = -216w^{-2}+2 \]We are finding critical points so we let \(dp/dw=0\)\[ 0= -216w^{-2}+2 \implies - 2 = -216w^{-2}\implies -2w^2=-216 \implies w^2=108 \]

OpenStudy (anonymous):

so the dimension would be 2 and 52

OpenStudy (anonymous):

We have a critical point when \(w=\pm \sqrt{108}\), we also have one when \(w=0\). However we know that \(w>0\), so the only critical point that matters is \(w=\sqrt{108}\).

OpenStudy (anonymous):

Now we just need to find the length. We go back to \(l=108/w = 108/\sqrt{108} = \sqrt{108}\).

OpenStudy (anonymous):

The dimensions are \(w=\sqrt{108}\) and \(l=\sqrt{108}\).

OpenStudy (anonymous):

wio you are a life saver you have time for one more ?

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