A rectangular piece of cardboard measuring 10in by 14 in is

A rectangular piece of cardboard measuring 10in by 14 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. Answer the following questions.

a) Give the restrictions on x.

< x <

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

V(x) =

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

= inches (Round to the nearest hundredth.)

d) What is this maximum volume?

= cubic inches (Round to the nearest hundredth.)

e) d) The volume will be greater than 24 cubic inches when

< x <

Solution

Length = 10
Width = 14

So, when we cut the square of edge, x we get :

length of box = 10 - 2x
width of box = 14 - 2x
height of box = x

Restrictions :
x > 0 of course
And x < 5 cuz thats the only way 10 -2x and 14-2x would be positive

So, 0 < x < 5

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V = x(14 - 2x)(10 - 2x)
V = x(140 + 4x^2 - 48x)
V = 4x^3 - 48x^2 + 140x ----> ANSWER

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c)
dV/dx = 12x^2 - 96x + 140 = 0

Pull a 4 out in common :

3x^2 - 24x + 35 = 0

Using the quadratic formula, we get :
x = 1.91833 , 6.08167

Obv, x = 6.08 is not possible as it
does not satisfy 0 < x< 5

So, x = 1.92 ----> ANSWER

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d)
MAx vol :
Plug in this x = 1.92 and find V

We get

Vmax = 120.16 ---> ANSWER

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e)
V > 24 :
Solving the inequality, we get :
4x^3 - 48x^2 + 140x > 24

4x^3 - 48x^2 + 140x - 24 > 0

Pull 4 out in common :
x^3 - 12x^2 + 35x - 6 > 0

0.18 < x < 4.47 ----> ANSWER