modeling with functions

Question

A box with an open top is to be constructed from a rectangular piece of cardboard with dimensions 12 in. by 20 in. by cutting out equal squares of side x at each corner and then folding up the sides 

a) Find the values of x for which the volume is greater than 230 in3. (Round your answers to three decimal places. Enter your answer using interval notation.)

Answer 1

The first big challenge here is setting up the function for Volume. By drawing a picture of the rectangle with the corners of length x cut out, we get a volume formula of:

V = (12-2x)*(20-2x)*x = 4x³-64x²+240x

The (12-2x) comes from the shorter side with two lengths of x cut out

The (20-2x) comes from the longer side with two lengths of x cut out

The factor of x comes from the size of the cuts giving the height of the box when the sides are folded up.

We then just need to set this function greater than or equal to 230:

4x³-64x²+240x >= 230

4x³-64x²+240x-230 >= 0

We can use a calculator or algebraic software like wolfram alpha to find the roots of the polynomial in the last step:

Approximately: x = 1.5067, 3.4584, 11.035

By looking at the graph, the part that we want is between the first two roots: [1.5067, 3.4584] and to the right of the third root: [11.035, infinity]

However, we have to keep in mind the physical limitations of the problem. Since the shorter side of the rectangle is 12, then the largest square we could cut out would have side length 6. This means we really just want the first interval above (not the one going to infinity)

Hope this is helpful!

Answer 2

Here's an interval rounded to 2 decimal places (1.51,3.46)

Largest volume is when x=2.42 giving volume = 262.7 cubic inches. 2.42 is about the midpoint between 1.51 and 3.46

x giving largest volume is found by taking the derivative of volume= V = x(12-2x)(20-2x) and setting it = 0

V' = 12x^2 - 128x + 240 =0

solve for x= 2.42 giving volume = 262.7