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

Answer:

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:

$V = l \times w \times h$ 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, $\frac{dV}{dx}$

V(x) = $20x^2-6x^3$        ......[4]

Now;

$\frac{dV}{dx} =40x - 18x^2$

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 $x = \frac{20}{9}$

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

$V(0) = 0$ and  

$V(\frac{20}{9} ) = 20(\frac{20}{9})^2-6(\frac{20}{9})^3$ = 32.92(apporx)

So,  V(x) = 32.92 which is maximum for $x = \frac{20}{9}$

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..