# Math Help - Dimensions of Dog Pen

1. ## Dimensions of Dog Pen

You wish to build a rectangular pen for your dog with 24 yards of fence. What are the dimensions (length and width) of the pen that will allow for maximum room to roam?

Is there a way to solve this question without using a graphing calculator or calculus?

Our teacher used a graphing calculator and found the maximum point to be (6,36) where the number 6 = the length and 36 = the width wirtten (L, W).

That's find and dandy but how does answer this question without a calculator and without calculus?

2. Originally Posted by magentarita
You wish to build a rectangular pen for your dog with 24 yards of fence. What are the dimensions (length and width) of the pen that will allow for maximum room to roam?

Is there a way to solve this question without using a graphing calculator or calculus?

Our teacher used a graphing calculator and found the maximum point to be (6,36) where the number 6 = the length and 36 = the width wirtten (L, W).

That's find and dandy but how does answer this question without a calculator and without calculus?
Yes there is another way to solve this by making perfect square (completing the square)

Let length of rectangular pen is x
and width of rectangular pen is y.

Perimeter $P = 2x + 2y.$

$\Rightarrow \; 24 = 2x + 2y$

$\Rightarrow \; 12 = x + y$

$\Rightarrow \; \boxed {\;y = 12 - x \;}$

Now Area $A= x \times y$

$A= x(12-x)$

$= -x^2+12x$

$A= -(x-6)^2+36$

So, the maximum area is 36 $m^2$ when x = 6 m, so that y = 12 - 6 = 6 m.

3. ## I knew it....

Originally Posted by Shyam
Yes there is another way to solve this by making perfect square (completing the square)

Let length of rectangular pen is x
and width of rectangular pen is y.

Perimeter $P = 2x + 2y.$

$\Rightarrow \; 24 = 2x + 2y$

$\Rightarrow \; 12 = x + y$

$\Rightarrow \; \boxed {\;y = 12 - x \;}$

Now Area $A= x \times y$

$A= x(12-x)$

$= -x^2+12x$

$A= -(x-6)^2+36$

So, the maximum area is 36 $m^2$ when x = 6 m, so that y = 12 - 6 = 6 m.
I knew there was another way.

Thanks