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 = $$Length \times Breadth = xy$$We are given that the area of the garden is 122 square feetSo, $$xy=122$$ ---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 = $$30x+20(\frac{122}{x})$$Total cost = $$\frac{2440}{x}+30x$$Now take the derivative to minimize the cost $$f(x)=\frac{2440}{x}+30x$$ $$f'(x)=-\frac{2440}{x^2}+30$$Equate it equal to 0 $$0=-\frac{2440}{x^2}+30$$ $$\frac{2440}{x^2}=30$$ $$\sqrt{\frac{2440}{30}}=x$$$$9.018 =x$$Now check whether it is minimum or not take second derivative $$f'(x)=-\frac{2440}{x^2}+30$$$$f''(x)=-(-2)\frac{2440}{x^3}$$Substitute the value of x$$f''(x)=-(-2)\frac{2440}{(9.018)^3}$$$$f''(x)=6.6540$$Since it is positive ,So the x is minimum Now find y Substitute the value of x in A$$(9.018)y=122$$ $$y=\frac{122}{9.018}$$ $$y=13.528$$ Hence the dimensions of the garden that minimize the cost is 9.018 feet(length) and 13.528 feet(width)