Question

Quadratic Functions:

1. You have 150 yards of fencing to enclose a rectangular region. One side of the rectangle does not need fencing. Find the dimensions of the rectangle that maximize the enclosed area. What is the maximum area?

 

Complete the following steps to solve the above problem:

a. Write the equation for the area of the rectangular region:

A =

b. Write the equation for the fencing required:

150 =

c. Solve the equation for fencing for .

d. Substitute the result of step c) into the area equation to obtain A as function of .

e. Write the function in the form of . f(x)=ax^2+bx+c

f. Calculate -b/2a If a < 0, the function has a maximum at this value.

This means that the area inside the fencing is maximized when = ?

g. Find the length of side .

h. Find the maximum area.

Answer 1

a. A=LW where L=length of rectangular area, W=width.

b. 150=2L+W.

c. W=150-2L. But the true perimeter of the area is P=2L+2W, so W=(1/2)(P-2L).

d. A=L(P-2L)/2=LP/2-L^2.

e. f(L)=-L^2+LP/2-A.

f. -b/2a=(-P/2)/(-2)=P/4. This is the vertex of f(L), and L=P/4 is the value for the maximum area.

g. From (c) W=(1/2)(P-2L)=P/4, so W=L=P/4. This makes the area a square of area P^2/16.

From (b) 150=2L+W=P/2+P/4=3P/4, making P=4/3*150=200, so W=L=200/4=50 yd.

h. Maximum area is 50^2=2500 sq yd.