Question

The length of an Algebra 2 textbook is 2 times the height. The sum of the length, width, and height of the box is 10 centimeters. a. Write the expressions for the dimensions of the book. b. Write a polynomial function for the volume of the book in factored form. c. Find the maximum volume of the book.

Answer 1

As per the given statement:

The length of an Algebra 2 textbook is 2 times the height.

Let height be x then;

⇒
`l = 2x` ......[1]; where l is the length.

Also, the sum of the length, width and height of the box is 10 cm.

⇒
`l+w+h=10` where w is the width. .,.....[2]

Substitute equation [1] in [2] we get;

`2x+w+x=10` oe

`3x+w =10` or

`w =10-3x` ......[3]

(a)

The dimensions of the box is :-

`l = 2x`

`w = 10 -3x`

`h =x`

(b)

Volume of the book is given by:

`V = l * w * h` where V is the volume.

Substitute equation [1] and [3] in above formula;

`V = (2x)(10-3x)(x)`

`V = (2x^2)(10-3x)`

The polynomial function for the volume of the book in the factored form:

`V(x)= (2x^2)(10-3x)`

(c)

To find the maximum volume of the book;

we would find the derivative of volume with respect to x i.e,
`(dV)/(dx)`

V(x) =
`20x^2-6x^3` ......[4]

Now;

`(dV)/(dx) =40x - 18x^2`

Set this derivative equal to 0.

`40x -18x^2 = 0` or

`2x(20-9x) =0`

By Zero Product Property states that if ab = 0, then

either a = 0 or b = 0, or we have both a and b are 0.

then we have;

`x = 0` and
`x = (20)/(9)`

Then substitute these values in equation [4] to get the values of V(x);

`V(0) = 0` and

`V((20)/(9) ) = 20((20)/(9))^2-6((20)/(9))^3` = 32.92(apporx)

So, V(x) = 32.92 which is maximum for
`x = (20)/(9)`

Therefore, the graph of function V(x) is shown below and we can clearly see that there is a maximum very close to 2.22..