Quadratic models in factor form

Question

width x

length 75-x

Max area x(75-x)

 

What is the width if we maximize the room. Calculatemat the width to maximize the room.

Answer 1

This problem becomes a negative parabola in which to determine the maximum area is to determine the vertex of the parabola. The way you could do this is by completing the square.

 

Let A(x)=x(75-x)=75x-x^2 = -x^2+75x = -(x^2-75x)

 

Divide the second term by 2, square it, and add it within the parathesis to get a perfect square.

 

A(x)=-(x^2-75x+(75/2)^2) + (75/2)^2

 

A(x)=-(x-75/2)^2+(75/2)^2

 

Hence, the maximum area is when x=75/2 = 37.5 and the maximum area is 75/2*75(1-1/2) = (75/2)^2

 

Thus, the maximum area is 1406.25 sq. units. Moreover the width is 37.5 units and the length is 75-75/2=75(1-1/2) = 75/2 = 37.5 units

 

Note: the length and width become equal in which becomes a square to give the maximum area.