Question

What are the dimensions of a rectangular wall that can be constructed with 400 feet of brick that will maximize the area enclosed by that brick

Answer 1

Answer:

The dimensions of the wall are 100 ft x 100 ft

Step-by-step explanation:

we know that

The perimeter of a rectangular wall is

$P=2(x+y)$

where

x is the length

y is the width

we have

$P=400\ ft$

so

$400=2(x+y)$

simplify

$200=x+y$

$y=200-x$ ----> equation A

The area of a rectangular wall is equal to

$A=xy$ ----> equation B

substitute equation A in equation B

$A=x(200-x)\\A=-x^2+200x$

This is the equation of a vertical parabola open downward (because the leading coefficient is negative)

The vertex represent a maximum

Convert the quadratic equation in vertex form

$A=-x^2+200x$

Factor -1

$A=-(x^2-200x)$

Complete the square

$A=-(x^2-200x+100^2)+100^2$

$A=-(x^2-200x+10,000)+10,000$

Rewrite as perfect squares

$A=-(x-100)^2+10,000$ ----> equation in vertex form

The vertex is the point (100,10,000)

The x-coordinate of the vertex represent the length of the wall for a maximum area

so

$x=100\ ft$

Find the value of y

equation A

$y=200-(100)=100\ ft$

therefore

The dimensions of the wall are 100 ft x 100 ft