A winter recreational rental company is fencing in a new storage area. They have two options. They can set it up at the back corner of the property and fence it in on four sides. Or, they can attach it to the back of their building and fence it in on three sides. The rental company has decided that the storage area needs to be 100 m2 if it is in the back corner or 98 m2 if it is attached to the back of the building. Determine the optimal design for each situation.

Rectangular area attached to the back of the building

two sides of legth 7 m and one side of 14 m

Step-by-step explanation:

We need to compare quantity of fencing material to be used in both cases

1.Option

A = 100 m²          dimensions of storage area  "x"    and "y"

x*y = 100     y = 100/x

The perimeter of the storage area is

p = 2*x + 2*y       ⇒ p = 2*x + 2*100/x

p(x) = 2*x + 200/x

Taking drivatives on both sides of the equation

p´(x) = 2 - 200/x²

p´(x) = 0         ⇒        2 - 200/x² = 0

2*x² - 200 = 0       x² = 100

x = 10 m

and  y = 100/10      

y = 10 m

Required fencing material in first option

2*10 + 2*10  =  40 m

2.-Option

Following the same procedure

A = 98 m²       y = A/x         y = 98/x

p = 2*x + y             p(x) =  2*x  +  98/x

p´(x)  = 2 - 98/x²             p ´(x) = 0

2 - 98/x²  = 0

2*x²  = 98           x² =  49

x = 7 m       and        y   =  98/ 7         y  = 14 m

Total quantity of fencing material

p = 2* 7 + 14         p = 28

Therefore option 2 is more convinient from economic point of view

Optimal design rectangular storage area with two sides of 7 m and one side of 14 m

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.