Thread: Optimization Problem

A container for wooden matches consists of an open-topped box(to contain the matches) that slides into an outer box, open at both ends. The length of the boxes is fixed at 10cm to match the length of the matches. The outer box is designed so that one side completely overlaps for gluing.

a) if the outer box is made from a sheet of cardboard that is 16cm by 10cm, what dimensions for the outer box will maximize the capacity?
b) What size shoudl the sheet of the cardboard be, to make the inner box in this case?

Originally Posted by pita

A container for wooden matches consists of an open-topped box(to contain the matches) that slides into an outer box, open at both ends. The length of the boxes is fixed at 10cm to match the length of the matches. The outer box is designed so that one side completely overlaps for gluing.

a) if the outer box is made from a sheet of cardboard that is 16cm by 10cm, what dimensions for the outer box will maximize the capacity?
b) What size shoudl the sheet of the cardboard be, to make the inner box in this case?

For a) let where x is the height of the box and y is the width. Then you want to maximize the volume of the box, . Solve the first equation for y in terms of x, substitute into the volume equation, and then set its derivative (in terms of x) equal to 0.

how did u get 3x + 2y=16, im not really understanding that

Originally Posted by pita

how did u get 3x + 2y=16, im not really understanding that

The cardboard sheet you use to make the matchbox must be 16cm long. When you fold the paper, you make four folds in it, and you would normally have just two sides of length x, but the problem specifies that you need an overlapping section, so there are three sides of length x because one is on top of the other.

thank you

so den for the inner box, would the equation be 2x + 2y=16 or jus one length