What dimensions will produce the greatest area for Sharon's puppy to play, if she purchased 40 feet of fencing to enclose three sides of a fence?

2 Answers
May 20, 2018

Answer:

If the shape is a rectangle, the area will be #200 sq ft#

Explanation:

The fencing is to be used for #3# sides, If we assume that the fourth side is a wall or an existing fence, then the shape is a rectangle.

Let the length of each of the shorter sides (the breadth) be #x#.
The length will be #40-2x#

#A = x(40-2x)#

#A= 40x-2x^2#

For a maximum, #(dA)/(dx) =0#

#(dA)/(dx) = 40-4x=0#

#" "x =10#

The dimensions will be #10 xx 20# feet, giving an area of #200sq ft.#

If the shape is to be an equilateral triangle:

#A = 1/2 ab sin60° = 1/2 xx40/3 xx40/3 xxsin60#

#A = 76.9 sq ft# which is much smaller than a rectangle.

If the fencing is used to form a semi-circle against a wall, the area will be:

#r = C/(2pi) = 80(2pi) = 12.732# feet

#A = pir^2 = 12.732^2 =162 sq ft#

May 20, 2018

Answer:

Using a quadratic to solve this question.

So the length of the side is #10" feet."#
So the length of the front is #40-2(10)=20" feet."#

The maximum area is #20xx10=200" feet"^2#

Explanation:

The wording: to enclose 3 sides of a fence implies there is at least one more side.

Assumption: The shape is that of a rectangle.

Set area as #A#
Set length of front as #F#
Set length of side as #S#

Given: #F+2S=40" "............................Equation(1)#

Known: #A=FxxS" "..............................Equation(2)#

From #Eqn(1)# we have #F=40-2S" "....Equation(1_a)#

Using #Eqn(1_a)# substitute for #F# in #Eqn(2)#

#color(green)(A=color(red)(F)xxS color(white)("dddd")->color(white)("dddd")A=color(red)((-2S+40))xxS)#

#color(green)(color(white)("ddddddddddddd")->color(white)("dddd")A = -2S^2+40S)#

This is a quadratic of general shape #nnn# as the squared term is negative. Thus there is a maximum value of #A# and it is at the vertex.

#color(brown)("A very useful trick to find the vertex")#

Using the beginnings of completing the square write as:

#A=-2(S^2color(red)(-40/2)S)#

#S_("vertex")=(-1/2)xxcolor(red)(-40/2)= +10#

So the length of the side is #10" feet."#
So the length of the front is #40-2(10)=20" feet."#

The maximum area is #20xx10=200" feet"^2#