Respuesta :
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 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