when food is packaged in cans, it is not necessary that the material in both ends and the side be the same cost. if the ends of the can cost E $/cm squared, the side costs S $/cm squared and the label (including glue to attach the label to the can) is L $/cm squared, show that the best shape (most economical) to build the can is when r= (S+L)h/2E, where r is the radius of the can and h is its height.
To determine the most economical shape of a can, we need to find the relationship between the variables involved - the cost of the ends (E $/cm^2), the cost of the sides (S $/cm^2), the cost of the label (L $/cm^2), the radius of the can (r), and the height of the can (h).
Let's start by determining the cost of the ends of the can. The surface area of both ends combined can be calculated as:
Surface Area of Ends = 2πr^2
Therefore, the cost of the ends of the can will be:
Cost of Ends = Surface Area of Ends * E
= (2πr^2) * E
Now, let's determine the cost of the sides of the can. The surface area of the side can be calculated as:
Surface Area of Side = 2πrh
Therefore, the cost of the sides of the can will be:
Cost of Sides = Surface Area of Side * S
= (2πrh) * S
Additionally, we need to consider the cost of the label attached to the can. The surface area of the label is the same as the surface area of the ends, as it wraps around the can. Therefore, the cost of the label will be:
Cost of Label = Surface Area of Ends * L
= (2πr^2) * L
To find the total cost of the can, we sum up the costs of the ends, sides, and the label:
Total Cost = Cost of Ends + Cost of Sides + Cost of Label
= (2πr^2) * E + (2πrh) * S + (2πr^2) * L
= 2πr^2(E + L) + 2πrhS
To simplify this equation, let's factor out 2πr:
Total Cost = 2πr(r(E + L) + hS)
Now, we want to minimize the total cost of the can to make it as economical as possible. To find the shape of the can that minimizes the total cost, we take the derivative of the total cost with respect to r and set it equal to zero:
d(Total Cost)/dr = 0
Differentiating the equation with respect to r, we get:
2π(r(E + L) + hS) + 2πr(E + L)(1) = 0
Simplifying, we have:
2πr(E + L + E + L) = - 2πhS
Dividing both sides by 2π, we get:
r(E + L + E + L) = - hS
Further simplifying:
r(2E + 2L) = - hS
Finally, isolating r, we divide both sides by 2(E + L):
r = - hS / (2(E + L))
Now, we can rearrange the equation to have a positive right-hand side by multiplying the numerator and denominator by -1:
r = hS / (2(E + L))
This matches the formula given in the question: r = (S + L)h / 2E.
Therefore, the best shape (most economical) to build a can is when the radius (r) is equal to (S + L)h / 2E.