What is the largest possible area for a rectangle of 42m perimeter but one side is a wall?

Question

Farmer Bob gets creative and decides to use the 42m of fencing material to build a pigpen against a straight wall!! This means the fencing only needs to be used for 3 sides!. Summarize your result - which dimensions gave the largest pigpen? What do you notice? Why?

Answer 1

P=2l+2w

42=l+2w

A=lw

42=l+2w-------> l=42-2w

substitute

A=(42-2w)(w)

0=42w-2w²

this is a parabola that opens downward so there is a maximum point, the vertex

the vertex is (h,k) where h is the "maximizing point" and k is the maximum area

h=-b/2a

h=-42/(-4)

h=10.5  feet

you have a square that is 10.5 by 10.5 but one side is not needed because of the wall

add the 10.5 feet that isn't needed to the opposite side to get 21 feet

you now have a rectangle that is 10.5 by 21 by 10.5 by "the wall"

the maximum area is...

A=l*w

A=21*10.5

A=220.5 square feet

notice that the rectangle is basically two squares side by side, each being 10.5 by 10.5

 

 

Answer 2

There are 2 possibilities because of the orientation of the pigpen.  You can have 2 lengths and 1 width, or 2 widths and 1 length.  The perimeter is the sum of the fencing sides.

 

Let the length = x

Let the width = y

 

For the first possibility:

 

2x + y = 42

 

 

For the second possibility:

 

x + 2y = 42

 

 

So now you have 2 equations to work with.

 

2x + y = 42                   eq1

  x + 2y = 42                 eq2

 

 

We can use the substitution method.  Substitute eq2 into eq1 to have eq1 in terms of y.

 

2(42 - 2y) + y = 42

 

 

Solve the equation for y.

 

84 - 4y + y = 42

 

-3y = -42

 

   y = 14

 

 

Substituting this value of y into eq2 to solve for x

 

x = 42 - 2(14)

x = 42 - 28

x = 14

 

 

So in both cases, the length and width are 14 ft. 

 

Largest possible area = 14 ft × 14 ft

                               = 196 ft²