You have just bought a new puppy and want to fence in an area in the backyard for her to roam. You buy 100 linear feet from Home Depot and have decided to make a rectangular fenced in area using the back of your house as a side. What are the dimensions to maximize the area your puppy can roam

50 ft * 25 ft

Step-by-step explanation:

Let the length of the rectangular area be x ft and the width of the rectangular area be y ft. You already have 100 ft of material for fencing. Also one side of your house is used in fencing.

The perimeter of the fence needed = x + 2y

Therefore:

100 = x + 2y

x = 100 - 2y

Also the area of the rectangular fence is:

Area (A) = length * breadth

A = xy

Substitute x = 100 - 2y

A = (100 - 2y)y

A = 100y - 2y²

Maximum area is at dA / dy = 0. Hence:

dA/dy = 100 - 4y

100 - 4y = 0

4y = 100

y = 25 ft

substitute value of y in x:

x = 100 - 2y = 100 - 2(25) = 100 - 50 = 50 ft

x = 50 ft

The dimension of the rectangular space is 50 ft * 25 ft

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

the answer is b i just did that

step-by-step explanation: