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?

Question

Ethan Adkins · Accepted Answer

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

Let length be #l# and width be #w# Perimeter = #180 ft. = 2(l+ w)#---------(1) and Area = #1800 ft.^2 = l xx w#----------(2) From (1), #2l+2w = 180# #=> 2l = 180-2w# #=> l = (180 - 2w)/2# #=> l = 90- w# Substitute this value of #l# in (2), # 1800 = (90-w) xx w # #=> 1800 = 90w - w^2# #=> w^2 -90w + 1800 = 0# Solving this quadratic equation we have : #=> w^2 -30w -60w + 1800 = 0# #=> w(w -30) -60 (w- 30) = 0# #=> (w-30)(w-60)= 0 # #therefore w= 30 or w=60# The possible widths of the play area are : 30 ft. or 60 ft.

Kennedy Adams · Answer

#30" or "60" feet"# #"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#

Jameson Allen · Answer

undefined The possible widths of the play area are $ 0 < w \leq \frac{180}{2} $, which simplifies to $ 0 < w \leq 90 $ feet.