Find the Reduis, height, and the minamum cost.

Question

A microwaveable cup-of-soup package needs to be constructed in the shape of cylinder to hold 200 cubic centimeters of soup. The sides and bottom of the container will be made of styrofoam costing 0.04 cents per square centimeter. The top will be made of glued paper, costing 0.07 cents per square centimeter. Find the dimensions for the package that will minimize production cost.

Answer 1

Any of these min/max calculus problems, well most, need a very carefully drawn figure with the important information shown.

PRIMARY EQUATION: This is an equation for the thing you want to minimize or maximize. In this case it is the cost. And you want to minimize it. So we will need an equations for Cost = ______________________

SECONDARY EQUATION, sometimes called the CONSTRAINT, is the information with the numbers. It is a way to put the primary equation into a single variable. And this one has a geometric shape (a cylinder) for which we know the formulas for Surface Area and Volume. It seems we may need both since the problem gives us constraints in both volume (must hold 200 cubic centimeters of soup) and because we are building the container, we will need the surface area (the area of all of the surfaces of the container).

Vcyl = (pi)(r^2)(h) pi is the constant 3.14159.... r is the radius, h is the height of the cylinder

Vcyl = 200 cubic centimeters so (pi)(r^2)(h) = 200 This is the contraint. It can be solved for r or for h, but h would be the best choice so that we do not need square roots.

200

h = ------------

(pi)(r^2)

This we will use in the primary equation (cost) to put it in a single variable, r.

Cost: Surface Area of a cylinder is 2(pi)(r)(h) for the "lateral" area = the sides of the cylinder, then the top and bottom are both circles so they each have an area of (pi)(r^2), but because the cost of the top is different than the cost of the bottom, we will need to deal with the top circle differently than we deal with the bottom circle.

We are told that the sides (lateral area) and the bottom (one of the circles) will be made of styrofoam costing $0.04 / sq cm

So the cost for the sides and bottom would become:

Cost = [ 2(pi)(r)(h) ] * 0.04 + [ (pi)(r^2) ] * 0.04 + still need to

area of sides * cost of side material area of bottom * cost of bottom material add cost of top

cost of the top is the area of the other circle multiplied by $0.07 / sq cm or [ (pi)(r^2) ] * 0.07

Putting these together, our PRIMARY EQUATION is:

Cost = [ 2(pi)(r)(h) ] * 0.04 + [ (pi)(r^2) ] * 0.04 + [ (pi)(r^2) ] * 0.07

combining like terms (the pi r^2 terms) our simplified PRIMARY EQUATION becomes

Cost = 0.04(2)(pi)(r)(h) + 0.11(pi)(r^2)

This we want in a single variable, so the expression we got for "h" when we rearranged the constraint expression, we will now substitute it here, into the primary equation so that it is only in "r"s.

Cost = 0.08(pi)(r)(200) 0.11(pi)(r^2)

------------------------ + -----------------------------

(pi)(r^2)

And this simplifies to: [ (pi)'s cancel, .08*200=16 top "r" cancels one of the bottom "r"s ]

Cost = 16/r + 0.11(pi)(r^2)

NOW - take derivative of Cost, set it equal to 0, find the "r" that minimizes the cost, plug it back in to the constraint equation to find the value for h. Then plug both "r" and "h" back into the primary equation to find the minimum cost. (I get answers with cube roots in them...)