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?
Here's a, b, and c .
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 .
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
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".