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12. An open top box is to be made by cutting congruent squares of side length x from the corners of a 20- by 25-inch               sheet of tin and bending the sides up.

a) Write an equatipon for the volume V as a function of x. What is the domain of this function?

b) What size corner squares should be cut to yield a box with a volume of 300 in3?

c) What size corner squares should be cut to yield a box with a volume more than 300 in3?

d) What size corner squares should be cut to yield a box with a of at most 300 in3?

e) How large should the square be to make the box be to make the box hold as much as possible? What is the resulting       volume?

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Here's  a,  b,  and  c .

a)

volume of box  =  (length )(  width )(height)

V            =  (25 - 2x)(20 - 2x)(x)

V            =  (500 - 90x + 4x2)(x)

V            =  500x - 90x2 + 4x3

V            =  4x3 - 90x2 + 500x

Since  x  is folded up from a side that is 20 inches long, x can't be larger than 10 inches. If x were 10 inches, it would fold 20 in half, and the width would be 0 .  So... the domain is all real  x | 0 ≤ x ≤ 10 .

b)

To find the side length of the squares (x) that yield a volume of 300 in3 ,

plug in  300  for  V and solve for  x .

300  =  4x3 - 90x2 + 500x

Here I used a graph to find the approximate solutions.

x ≈ 0.68 inches  and  x ≈ 7.93 inches  are the only values in the domain.

So...the size of the corner squares can be....

≈ 0.68 by 0.68    or     ≈ 7.93 by 7.93

c)

We can look at the graph again to see that the x values that cause a volume bigger than 300 are those between 0.68 and 7.93 , and those greater than 13.89 .

Since those greater than 13.89 are outside the domain, the x values that cause the volume to be bigger than 300 are......          0.68 < x < 7.93 .

So....for instance, the size of the corner square could be  0.5"  by  0.5",  or  1"  by  1", or  3"  by  3".

hectictar  Sep 12, 2017

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