find the dimensions of a rectangle with an area of 108 square feet that has the minimum perimeter?
Do you know the formula for area and perimeter?
a=lw and perimeter 2l+2w ?
yeah
We want to minimize perimeter, so we can use the area formula to find \(w\) in terms of \(l\).
Remember that they told us \(a=108\).
okay that's where I get lost idk whats after that
Okay so \[ 108 = lw\\ p = 2l+2w \]
Do you see how we can get rid of \(l\) by using substitution?
not quite how would It be done
\[ l = 108/w \]
that's all
Now substitute: \[ p=2(108/w)+2w \]
In order to minimize \(p\) you need to find \(dp/dw\). It will tell you which width \(p\) has max/min at.
You need to find \(dp/dw = 0\)
wio your a genius how do I find dp/dw or is it 0 then where do I plug it in
First you differentiate \(p\) in terms of \(w\).
Can you differentiate: \[ p=2(108/w)+2w \]?
p-2w=2(108/w) ?
Find the derivative.
Do you know how to differentiate?
\[ p = 216w^{-1}+2w \]
Just use power rule.
then would it ;be p= 216/w2+2w =0?
You don't know how to differentiate?
not really
\[ \frac{d(216w^{-1})}{dw} = -216w^{-2} \]
\[ \frac{d(2w)}{dw} = 2 \]
This gives us: \[ \frac{dp}{dw} = -216w^{-2}+2 \]We are finding critical points so we let \(dp/dw=0\)\[ 0= -216w^{-2}+2 \implies - 2 = -216w^{-2}\implies -2w^2=-216 \implies w^2=108 \]
so the dimension would be 2 and 52
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 \(w=\sqrt{108}\).
Now we just need to find the length. We go back to \(l=108/w = 108/\sqrt{108} = \sqrt{108}\).
The dimensions are \(w=\sqrt{108}\) and \(l=\sqrt{108}\).
wio you are a life saver you have time for one more ?
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