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dimensions to minimize cost of a soup can ?

hello i was wondering if someone could help me solve this problem:

Suppose that you own a company that produces cans of soup. A 10 ¾ ounce can of soup is to
contain 12 cubic inches of soup. The cost of materials for the top and bottom of the can is 5
cents per square inch, and the cost of the materials for the side of the can is 3 cents per square
inch. To minimize cost, what should the dimensions of the can be?

1. Solve the problem to determine the proper dimensions to minimize the cost of
manufacturing each can. What is the minimum cost of the material? (3 points)

2. Suppose that the ingredients of the soup costs $.95 for each can of soup that is
produced, and the minimum cost of the materials to make each can is the amount you
derived in part 1 above, and when the fixed cost is factored in (the factory, employee
salaries, management salaries, utilities, etc.), the fixed cost is $1500 per day. Also,
suppose that you sell each can of soup for $3.79. Determine the (a) overall cost
function of producing the cans each day, (b) the revenue function, and (c) the profit
function. How many cans does your company need to produce each day to start making
a profit?

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Francisco E. | Francisco; Civil Engineering, Math., Science, Spanish, Computers.Francisco; Civil Engineering, Math., Sci...
5.0 5.0 (1 lesson ratings) (1)
The target equation for question 1 is
The constraint is
∏xR2Xh = 12 cubic inches

Objective Cell (Min)
Cell Name Original Value Final Value
$F$3 value 0.502656          1.03
Variable Cells
Cell Name Original Value Final Value Integer
$F$4 r 1 1.05 in Contin
$F$5 h 1 3.49 inContin

Cell Name Cell Value Formula Status Slack
$F$6 volume 12 $F$6=12 Binding 0
The minimum cost of the can will be 1.03 dollars, the dimensions will be: base radius 1.05 inches, height 3.49 inches and the volume is 12 cubic inches. 
Part two has the following:
ingredients 0.95
can 1.03
a) overall daily cost will be (1.03+0.95)x + (1500/x)=, being x the number of cans produced daily.
b) Revenue function will be 3.79X being X the same number of cans produced daily.
c) Profit will be b-c or (3.79x)-((1.03+0.95)x + (1500/x))
For the last question, make the profit equal to zero and solve for x. It will be the break even point.