Jessica R.

asked • 02/20/21# MATH HELP PLZ!!!

A rug cleaning company sells three models. EZ model weighs 10 pounds, packed in a 10-cubic-foot box. Mini model weighs 20 pounds, packed in an 8-cubic-foot box. Hefty model weighs 60 pounds, packed in a 28-cubic-foot box. A delivery van has 284 cubic feet of space and can hold a maximum of 440 pounds. To be fully loaded, how many of each should it carry if the driver wants the maximum number of Hefty models?

## 1 Expert Answer

Daniel B. answered • 02/21/21

A retired computer professional to teach math, physics

Let

e be the number of EZ models,

m be the number mini models,

h be the number of hefty models.

In the optimal solution all three variables have positive integer values and satisfy the equations:

10e + 20m + 60h = 440 (expresses that the van is fully loaded in the sense of weight)

10e + 8m + 28h = 284 (expresses that the van is fully loaded in the sense of volume)

Use the first equation to express e in terms of m and h:

e = 44 - 2m - 6h

Subtract the two equations:

12m + 32h = 156

Divide the above equation by 4:

3m + 8h = 39

From the above, express m in terms of h:

m = (39 - 8h)/3

Now we look for integer solutions for the above equation.

Given that we want to maximize h, we can look for integer solutions starting with the

largest possible h, which is 4.

But h = 4 would give non-integer m.

The next possibility is h = 3, which does yield an integer value for m.

So

h = 3,

m = 5,

e = 16

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Mark M.

This problem assumes (erroneously) that the boxes are mailable and can be "squeezed" into the given volume.02/21/21