A soda is to be sold in hexagonally based cans, each holding 400 milliliters of

Question

A soda is to be sold in hexagonally based cans, each holding 400 milliliters of pop. The material for the sides costs 0.03

cents per square centimeter. The material for the bottom costs 0.05 cents per square centimeter. The material for the top costs 0.07 cents per square centimeter. What are the optimal dimensions for the

can (height and side length)?

So far I have the following:

Volume = 400 mL

Volume (can) = Area (Hex) * h

Surface Area (can) = 2* A (Hex) + 6r*h

Cost --> C = 0.05A + 0.07A + 0.03 *6r*h

                C= 0.012 A + 0.18r*h

Since we need everything in terms of h,

Area (Hex) = 400 mL/h

The new cost equation: C = 0.012(400/h) + 0.18 r*h

C = 48/h + 0.18 r*h

There are two many variables, so I am stuck in finding out a way to get rid of the radius, r. in the cost equation. My teacher said to use the area of a polygon to find r where A = 1/2 *n*r^2*sin(2pi/n) given n = 6 (sides). How do you solve for r??

After finding the cost equation and taking the derivative, d/dh must be found to find the critical values. Finally, find the

I would appreciate an explanation. Thank you.

Explanation / Answer

A = 1/2 *n*r^2*sin(2pi/n) and A = 400/h

equating the two we get

1/2 *6*r^2*sin(2pi/6) = 400/h

==> 3*sqrt(3)*r^2/2 = 400/h

==> r = 20*sqrt(2/(h*3*sqrt(3)))

now C = 48/h + 0.18 r*h

==> C = 48/h + 0.18* 20*sqrt(2/(h*3*sqrt(3)))*h

==> C = 48/h + 3.6*sqrt(2/(3*sqrt(3))*sqrt(h)

dC/dh = -48/h^2 + 3.6*sqrt(2/(3*sqrt(3))/2sqrt(h) = 0

==> 48/h^2 = 1.58/sqrt(h)

==>48*sqrt(h) = 1.58*h^2

==> (48*sqrt(h))^2 = (1.58*h^2)^2

==> 2304*h = 2.4964*h^4

==> h^3 = 922.92 ==> h = 9.736

just put the value of h in the equation and calculate and u will get the value of r

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