Stanford university's soccer field had an area of 8800 square yards. it's length is 30 yards longer than its width. suppose a groundskeeper decides that he had enough chalk to line 600 feet of the perimeter of the field. what is the maximum area of the field this chalk could outline assuming the length remains 30 yards longer than its width

2275 square yards

Step-by-step explanation:

1 yard = 3 feet

We know that:

L = W + 30

where:

L = length

W = Width

We know that the soocer field has an area of 8800 square yards and that the equation to get this is:

L * W = A

(W+30) * W = 8800

W^2 + 30 W = 8800

W^2 + 30W - 8800 = 0

(W+110)*(W-80) = 0

W=80

W=-110

W=80 yards and L = 80 + 30 = 110 yards This is the lenght of all the field.

The problem is asking us what is the maximum area that a ground keeper can outline assuming the the lenght remains 30 yards longer than the width

He has chalk to line 600 feet = 200 yards

Now we can assume that:

2L + 2W = 200 yards

2(W+30) + 2W = 200 yards

2W + 60 + 2W = 200

4W = 140

W = 140/4

W = 35 yards

L= 35 + 30 = 65 yards

To know the maximum area we have to  multiply:

35*65=2275 square yards

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point