MATH SOLVE

3 months ago

Q:
# A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft and on the other three sides by a metal fence costing $10/ft. If the area of the garden is 122 square feet, find the dimensions of the garden that minimize the cost.

Accepted Solution

A:

Answer:The dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)Step-by-step explanation:Let the length of garden be xLet the breadth of garden be y Area of Rectangular garden = [tex]Length \times Breadth = xy[/tex]We are given that the area of the garden is 122 square feetSo, [tex]xy=122[/tex] ---AA landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $20/ft So, cost of brick along length x = 20 xOn the other three sides by a metal fence costing $10/ft. So, Other three side s = x+2ySo, cost of brick along the other three sides= 10(x+2y)So, Total cost = 20x+10(x+2y)=20x+10x+20y=30x+20yTotal cost = 30x+20ySubstitute the value of y from ATotal cost = [tex]30x+20(\frac{122}{x})[/tex]Total cost = [tex]\frac{2440}{x}+30x[/tex]Now take the derivative to minimize the cost [tex]f(x)=\frac{2440}{x}+30x[/tex] [tex]f'(x)=-\frac{2440}{x^2}+30[/tex]Equate it equal to 0 [tex]0=-\frac{2440}{x^2}+30[/tex] [tex]\frac{2440}{x^2}=30[/tex] [tex]\sqrt{\frac{2440}{30}}=x[/tex][tex]9.018 =x[/tex]Now check whether it is minimum or not take second derivative [tex]f'(x)=-\frac{2440}{x^2}+30[/tex][tex]f''(x)=-(-2)\frac{2440}{x^3}[/tex]Substitute the value of x[tex]f''(x)=-(-2)\frac{2440}{(9.018)^3}[/tex][tex]f''(x)=6.6540[/tex]Since it is positive ,So the x is minimum Now find y Substitute the value of x in A[tex](9.018)y=122[/tex] [tex]y=\frac{122}{9.018}[/tex] [tex]y=13.528[/tex] Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)