Question

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?

Answer 1

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`

Answer 2

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`