   Chapter 10.4, Problem 28E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# Volume(a) A square piece of cardboard 36 centimeters on a side is to be formed into a rectangular box by cutting squares with length x from each corner and folding up the sides. What is the maximum volume possible for the box?(b) Show that if the piece of cardboard is k centimeters on each side, cutting squares of size k/6 and folding up the sides gives the maximum volume.

(a)

To determine

To calculate: The maximum volume of the box.

Explanation

Given Information:

x is the side of a square piece cut from each corner of a square with 36 cm sides to convert it into a rectangular box.

Formula used:

To find the maximum value, calculate the relevant stationary value of an equation. Differentiate the function with respect to the independent variable and equate it to 0. the relevant value is the stationary value that satisfies the provided conditions.

If the second derivative of the provided function is less than zero, then substituting the value of the independent variable will give the maximum value of the equation.

The product rule is used when the function is a product of two differentiable terms. If p(x)=f(x)g(x), then the derivative is

p'(x)=f'(x)g(x)+f(x)g'(x)

Volume of a rectangular box is given by:

V=lbh

Where, l is length, b is breadth and h is height of the box.

Calculation:

As given, x is the side of a square piece cut from each corner of a square with 36 cm sides to convert it into a rectangular box.

Thus, two lengths are cut from each corner, that is 2x from length and 2x from breadth if x is height of the box.

The volume of the box is

V=lbhV=(362x)(362x)x

To find the stationary value for V=(362x)(362x)x, differentiate the equation with respect to the independent variable x to get

V'=(362x)(364x)+(2)(36x2x2)=3626×36x+8x22×36x+4x2=12x28×36x+362=12(x224x+108)

Put V'=0 to get x,

12(x224x+108)=0x224x+108<

(b)

To determine

To prove: Cutting squares of side k6 when piece of cardboard is k cm gives the maximum volume of rectangular box.

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