1. ## Optimization Problem

The question is:
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 $\displaystyle y= 300-3/2x$. I have also solved the triangle on top of the rectangle where the $\displaystyle h= x(3)^(0.5)$. (its x times root 3)
I am just not sure what to do from there.

2. ## Ok

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

$\displaystyle 600=3x+2y$....and you have that $\displaystyle A=xy+x^2$...now solve for a variable imput it into the area equation differentiate them find a max

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

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

4. ## You must have made a mistake

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

I get $\displaystyle A(126.8) = 20884.7cm$ but the answer is $\displaystyle 2.1 m^2$
$\displaystyle 600=3x+2y$...so $\displaystyle y=\frac{-3(x-200)}{2}$...imputting that into your area equation you get $\displaystyle A=x\cdot\bigg(\frac{-3(x-200)}{2}\bigg)$..differentiating we get $\displaystyle A'=300-3x$ which equals zero at x is equal to 100...now to check if its a max we find the second derivative $\displaystyle A''=-3$...which is always negative...therefore $\displaystyle x=100$ is a max...substitute it into your perimeter equation and get x

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

6. ## Of course

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?

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