Question

A rectangular piece of cardboard measuring

16


in by

24


in is to be made into a box by cutting equal size squares from each corner and folding up the sides. Let x represent the length of a side of each such square in inches. Complete parts ​(a) through ​(d) below.

(a) Give the restrictions on x.

​(b) Determine a function V that gives the volume of the box as a function of x.

(Round to the nearest​ hundredth.)

​(c) For what value of x will the volume be a​ maximum?

(Round to the nearest​ hundredth.)

(d)The volume will be greater than

288 cubic inches when

(Type an integer or a decimal. Round to the nearest​ thousandth.)

Answer 1

Answer:

(a) x<8

(b) $\displaystyle v=384x-80x^2+4x^3$

(c) $\displaystyle x=3.13$

(d) $\displaystyle 0.917<x<6$

Step-by-step explanation:

Optimization

It is the procedure to find the set of values for the variables of a function such that it reaches a maximum or a minimum value. If equalities are given as relationships between the variables, then the derivative is a suitable method to find the critical points or candidates for extrema values.

The problem at hand is about a geometric maximization, given some dimensional conditions. First, we have a rectangular piece of cardboard measuring 24 x 16 inches. A box is to be made out of that cardboard by cutting equal size squares from each corner and folding up the sides of length x.  

(a)

$\displaystyle w=24\ in$

$\displaystyle L=16\ in$

When we do so, the base of the box will have dimensions

$\displaystyle W'=24-2x$

$\displaystyle L'=16-2x$

Since the new width of the base must be positive, then

$\displaystyle 24-2x>0$

which poses the restriction

$\displaystyle x<12$

The same situation happens with the length

$\displaystyle 16-2x>0$

x<8

Since this last condition is more restrictive than the first, we state that x must be less than 8

(b) The volume of the box is the product of the area of the base by the height x

$\displaystyle v=L'.W'.x$

$\displaystyle v=(16-2x)(24-2x)\ x$

Operating

$\displaystyle v=(384-80x+4x^2)x$

$\displaystyle v=384x-80x^2+4x^3$

(c)

To find the maximum volume, we take the first derivative of V:

$\displaystyle v'=384-160x+12x^2$

Equating to zero to find the critical points

$\displaystyle v'=0$

$\displaystyle 12x^2-160x+384=0$

Dividing by 4:

$\displaystyle 3x^2-40x+96=0$

The roots of the equation are:

$\displaystyle x=10.19,x=3.13$

Since x=10.19 is out of the restrictions found in part a, the only valid solution is

$\displaystyle x=3.13$

We must test if the critical point is a maximum or a minimum, by computing the second derivative

$\displaystyle v''(x)=-160+24x$

$\displaystyle v''(3.13)<0$

Since the second derivative is negative, the value is a maximum

(d) We must find all the values of x such as v>288:

$\displaystyle 384x-80x^2+4x^3>288$

Rearranging

$\displaystyle 4x^3-80x^2+384x-288>0$

Simplifying by 4

$\displaystyle x^3-20x^2+96x-72>0$

Factoring

$\displaystyle (x-6)(x^2-14x+12)>0$

$\displaystyle (x-6)(x-13.083)(x-0.917)>0$

We found three real and positive roots for the third-degree polynomial

$x=6,\ x=0.917,\ x=13.083$

The function is positive when

$\displaystyle 0.917<x<6$

or

$\displaystyle x>13.083$

The only interval lying into the valid values of x is

$\displaystyle 0.917<x<6$