1. ## Length and breadth of rectangle from perimeter

Hi,

Looking for a hint on this one. Right now a bit clueless on how to proceed.

ABCD is a rectangular piece of land and XY is a wall. A fence 60 meters long is used to fence up the land. Find the length and the breadth of the rectangular piece of land when the area is maximum.

I denoted AB = p and BC = q.

Given,

$\displaystyle \begin{tabular}{ r l } $$2p + q$$ &= $$60$$ \\ $$q$$ &= $$60 - 2p$$ \\ &= $$2(30 - p)$$ \\ \\ $$Area$$ &= $$p \times q$$ \\ \ &= $$2p(30 - p)$$ \\ \end{tabular}$

Am I on the right track. I don't see any other data to use here. Can you please provide a hint on how to proceed.

Thanks!

2. Originally Posted by mathguy80
Hi,

Looking for a hint on this one. Right now a bit clueless on how to proceed.

ABCD is a rectangular piece of land and XY is a wall. A fence 60 meters long is used to fence up the land. Find the length and the breadth of the rectangular piece of land when the area is maximum.

I denoted AB = p and BC = q.

Given,

$\displaystyle \begin{tabular}{ r l } $$2p + q$$ &= $$60$$ \\ $$q$$ &= $$60 - 2p$$ \\ &= $$2(30 - p)$$ \\ \\ $$Area$$ &= $$p \times q$$ \\ \ &= $$2p(30 - p)$$ \\ \end{tabular}$

Am I on the right track. I don't see any other data to use here. Can you please provide a hint on how to proceed.

Thanks!
You are right where you need to be!

The quadratic $\displaystyle A(p)=2p(30-p)=-2p^2+60p$ maximum can be found in a few different ways.

1st symmetry the vertex must be at the midpoint of the p-intercepts. Since $\displaystyle p=0,p=30$ the midpoint is at $\displaystyle p=15$ this gives

$\displaystyle A(15)=2(15)^2=450$

or you can expand this out and use the vertex formula$\displaystyle \displaystyle p=\frac{-b}{2a}=\frac{-60}{-4}=15$

3. Thanks @TheEmptySet. Guess I blanked out at the final dash. I didn't simplify the expression and just didn't think the problem through. And didn't know about the vertex formula. You explained it really well. Thanks a lot. Learned some great new stuff today!