# Optimization Problem

• April 8th 2008, 04:31 PM
rhhs11
Optimization Problem
The question is:
Quote:

A gable window has the form of a rectangle topped by an equilateral triangle, the side of which are equal to the width of the rectangle. Find the maximum area of the window if the perimeter is 600cm.
What i have done so far is that i have solved for y where $y= 300-3/2x$. I have also solved the triangle on top of the rectangle where the $h= x(3)^(0.5)$. (its x times root 3)
I am just not sure what to do from there.

• April 8th 2008, 04:41 PM
Mathstud28
Ok
Quote:

Originally Posted by rhhs11
The question is:
What i have done so far is that i have solved for y where $y= 300-3/2x$. I have also solved the triangle on top of the rectangle where the $h= x(3)^(0.5)$. (its x times root 3)
I am just not sure what to do from there.

$600=3x+2y$....and you have that $A=xy+x^2$...now solve for a variable imput it into the area equation differentiate them find a max
• April 8th 2008, 04:55 PM
rhhs11
ok, so i differentiated and i got $x=600/(3+(3)^(0.5)$ which equals 126.8. So i subbed that into total Area equation which is $A=300x - 3x^2 + x^2(3)^(0.5)/x$.

I get $A(126.8) = 20884.7cm$ but the answer is $2.1 m^2$ (Worried)
• April 8th 2008, 05:12 PM
Mathstud28
You must have made a mistake
Quote:

Originally Posted by rhhs11
ok, so i differentiated and i got $x=600/(3+(3)^(0.5)$ which equals 126.8. So i subbed that into total Area equation which is $A=300x - 3x^2 + x^2(3)^(0.5)/x$.

I get $A(126.8) = 20884.7cm$ but the answer is $2.1 m^2$ (Worried)

$600=3x+2y$...so $y=\frac{-3(x-200)}{2}$...imputting that into your area equation you get $A=x\cdot\bigg(\frac{-3(x-200)}{2}\bigg)$..differentiating we get $A'=300-3x$ which equals zero at x is equal to 100...now to check if its a max we find the second derivative $A''=-3$...which is always negative...therefore $x=100$ is a max...substitute it into your perimeter equation and get x
• April 8th 2008, 05:43 PM
rhhs11
im sorry, but i have been stuck on the same question for almost 2 hrs now. Can you please expand the differentiation?

(Worried)
• April 8th 2008, 05:55 PM
Mathstud28
Of course
Quote:

Originally Posted by rhhs11
im sorry, but i have been stuck on the same question for almost 2 hrs now. Can you please expand the differentiation?

(Worried)

Of course you can this is the point of the site my friend :)..ok $A=x\cdot\bigg(\frac{-3(x-200)}{2}\bigg)=x\cdot\bigg(\frac{-3x+600}{2}=\frac{-3x^2}{2}+300x\bigg)$...so using basic rules of diffentiation we get that $A'=2\cdot\frac{-3x^{2-1}}{2}+1\cdot{300x^{1-1}}=-3x+300x^0=-3x+300=300-3x=3(100-x)$