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.

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 - 2x

The area of rectangle is = length×width

The area of rectangle is = (x)×(60 - 2x)

A = 60x - 2x²

The 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/2a

In the above function a = -2 and b = 60

Substitute the value of a and b in x max = -b/2a

x max = -60/-4 = 15 feet

Thus the value of x is 15, now calculate the value of y as shown:

y = 60 - 2x

y = 60 - 30

y = 30

Hence, the dimensions of the rectangular area are 15 ft and 30 ft.

32/80  

91/140

the last one 54 doesn't go into 45. 54/100 ?

step-by-step explanation: