What is the maximum volume in cubic inches of an open box to be made from a 12-inch by 16-inch piece of cardboard by cutting out squares of equal sides from the four corners and bending up the sides? your work must include a statement of the function and its derivative. give one decimal place in your final answer.

step-by-step explanation:

the volume of a box is: v = l * w * h. we are starting with a piece of cardboard 10 inches by 20 inches. if we let x represent the height of the box, x will also be the side of the square taken out of each corner.

the length of the box becomes 20 - 2x. the width of the box becomes 10 - 2x. the height of the box is x.

substituting in the equation for the volume gives:

v = x * (20 - 2x) * (10 - 2x)

multiply the right side to get:

v(x) = 4x3 - 60x2 + 200x

take the first derivative of v:

v'(x) = 12x2 - 120x + 200

to find x, set the first derivative equal to 0:

12x2 - 120x + 200 = 0

use the quadratic formula to solve for x:

a = 12

b = -120

c = 200

x = (120 ± √)2 - 4 * 12 * 200) / (2 * 12)

use the positive value first:

x = (120 + √(14400 - 9600) / 24

x = (120 + 69.3) / 24

x = 189.3 / 24

x = 7.9 inches

this solution can be discarded because it would make the width a negative number.

use the negative value next:

x = (120 - √(14400 - 9600) / 24

x = (120 - 69.3) / 24

x = 50.7 / 24

x = 2.1 inches

take the second derivative of v to check for a local maximum.

v''(x) = 24x - 120

substitute for x:

v''(2.1) = 24 * 2.1 - 120

the second derivative is less than zero, x is a maximum.

sides of the box:

length = 20 - 2 * 2.1 = 17.8 inches

width = 10 - 2 * 2.1 = 7.8 inches

height = 2.1 inches

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

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step-by-step explanation: