A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. He will use one side of his house as a side of the play area. What dimensions should he use if he wants to maximize the play area?

1 Answer
Jun 11, 2017

Answer:

Maximum area is #450# square feet when dimensions of play area are #30'xx15'#, where#30# feet is along the side of the house.

Explanation:

Let #l# be the length along the side of the house and #w# be the width.

Hence fencing required will be #l+w+w=l+2w# and this is #60#feet. In other words #l+2w=60# i.e. #w=(60-l)/2=30-l/2#

Area covered by this will be #lxx(30-l/2)=30l-l^2/2#

= #-1/2(l^2-60l)#

= #-1/2(l^2-60l+900)+450#

= #-1/2(l-30)^2+450#

It is apparent that as coefficient of #(l-30)^2# is #-1/2#,

#-1/2(i-30)^2# is alwaays negative, except that it is #0# when #l=30# and hence maximum area at this level is #450# square feet and dimensions of play area will be #30'xx15'#, where#30# feet is along the side of the house.