A cylinder shaped can needs to be constructed to hold 550 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.07 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

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Step-by-step explanation:

The cost of the ends of the can will be ...

  c1 = 0.07(2πr²)

The cost of the side of the can will be ...

  c2 = 0.04(2πrh)

The volume of the can will be ...

  v = πr²h

We want the derivative of the total cost to be zero, and we want the volume to be 550 cm³. We can take the derivatives of both equations to find a relation between r and h.

  d(c1 +c2) = 0.28πr·dr +0.08π(h·dr +r·dh) = 0

  d(v) = 2πrh·dr +πr²·dh = 0

Solving the first equation for dh/dr gives ...

  dh/dr = -π(0.28r +0.08h)/(π(0.08r)) = -(7r+2h)/(2r)

Solving the second equation for dh/dr gives ...

  dh/dr = -2πrh/(πr²) = -2h/r

Equating these expressions, we get ...

  -(7r +2h)/(2r) = -2h/r

  7r +2h = 4h . . . . . . . multiply by -2r

  h = 7/2r . . . . . . . . . . subtract 2h, divide by 2

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Now, we can find the can dimensions from the volume equation.

  550 = πr²(7/2r)

  r³ = 1100/(7π)

  r ≈ ∛50.02 ≈ 3.685 . . . . . cm

  h = 7/2(3.685 cm) = 12.896 cm

The can cost will be a minimum when the radius is 3.685 cm and the height is 12.896 cm.

Additional comment

You may notice that the ratio of height to diameter is the same as the ratio of end cost to side cost: 7/4. This is the generic solution to this sort of problem.

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>