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 in^{3}?

c) What size corner squares should be cut to yield a box with a volume more than 300 in^{3}?

d) What size corner squares should be cut to yield a box with a of at most 300 in^{3}?

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?

AdamTaurus Sep 12, 2017

#1**+1 **

Here's a, b, and c .

a)

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

V = (25 - 2x)(20 - 2x)(x)

V = (500 - 90x + 4x^{2})(x)

V = 500x - 90x^{2} + 4x^{3}

V = 4x^{3} - 90x^{2} + 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 in^{3} ,

plug in 300 for V and solve for x .

300 = 4x^{3} - 90x^{2} + 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