Vanessa has 180 feet of fencing that she intends to use to build a rectangular play area for her dog. She wants the play area to enclose at least 1800 square feet. What are the possible widths of the play area?

2 Answers
Nov 12, 2017

The possible widths of the play area are : 30 ft. or 60 ft.

Explanation:

Let length be ll and width be ww

Perimeter = 180 ft. = 2(l+ w)180ft.=2(l+w)---------(1)

and

Area = 1800 ft.^2 = l xx w1800ft.2=l×w----------(2)

From (1),

2l+2w = 1802l+2w=180

=> 2l = 180-2w2l=1802w

=> l = (180 - 2w)/2l=1802w2

=> l = 90- wl=90w

Substitute this value of ll in (2),

1800 = (90-w) xx w 1800=(90w)×w

=> 1800 = 90w - w^21800=90ww2

=> w^2 -90w + 1800 = 0w290w+1800=0

Solving this quadratic equation we have :

=> w^2 -30w -60w + 1800 = 0w230w60w+1800=0

=> w(w -30) -60 (w- 30) = 0w(w30)60(w30)=0

=> (w-30)(w-60)= 0 (w30)(w60)=0

therefore w= 30 or w=60

The possible widths of the play area are : 30 ft. or 60 ft.

Nov 12, 2017

30" or "60" feet"

Explanation:

"using the following formulae related to rectangles"

"where "l" is the length and "w" the width"

• " perimeter (P) "=2l+2w

• " area (A) "=lxxw=lw

"the perimeter will be "180" feet "larrcolor(blue)"fencing"

"obtaining "l" in terms of "w

rArr2l+2w=180

rArr2l=180-2w

rArrl=1/2(180-2w)=90-w

A=lw=w(90-w)=1800

rArrw^2-90w+1800=0larrcolor(blue)"quadratic equation"

"the factors of + 1800 which sum to - 90 are - 30 and - 60"

rArr(w-30)(w-60)=0

"equate each factor to zero and solve for "w

w-30=0rArrw=30

w-60=0rArrw=60