Answer:
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 Â Â 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   Â
y = 10 m
Required fencing material in first option
2*10 + 2*10 Â = Â 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 Â Â Â Â 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