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

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

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