MATH SOLVE

5 months ago

Q:
# Cesar wants to fence three sides of a rectangular exercise yard for his dog. The fourth side of the exercise yard will be a side of the house. He has 60 feet of fencing available. Find the dimensions that will enclose the maximum area. The fence parallel to the house is feet, the fence perpendicular to the house is feet and the area of the yard is square feet.

Accepted Solution

A:

Answer:The dimensions of the rectangular area are 15 ft and 30 ft.Step-by-step explanation:Consider the provided information.Cesar wants to fence three sides of a rectangular exercise yard for his dog. The fourth side of the exercise yard will be a side of the house. He has 60 feet of fencing available. We need to find the dimensions that will enclose the maximum area. Let the length of the fence is x feet.Let the width of the fence is y feet.The total fencing available is 60 feet.Thus, width is: y = 60 - x - x = 60 - 2xThe area of rectangle is = length×widthThe area of rectangle is = (x)×(60 - 2x)A = 60x - 2x²The above function opens downwards as the coefficient of x² is a negative number, thus the maximum of the function can be calculated as:x max = -b/2aIn the above function a = -2 and b = 60Substitute the value of a and b in x max = -b/2ax max = -60/-4 = 15 feet Thus the value of x is 15, now calculate the value of y as shown:y = 60 - 2xy = 60 - 30y = 30Hence, the dimensions of the rectangular area are 15 ft and 30 ft.