Find the most economical dimensions for a closed cylindrical can containing a quart?

For your first question that you put as title,

Let the two numbers be represented by a and b.
Their product is 64, so:

ab = 64

Their sum is a minimum.

So, the sum S of a and b, S = a + b should be minimum.

To get the minimum, you have to find the derivative of S (or find the shape of the graph of S as a and b vary.

S = a + b

Substitute either a = 64/b or b = 64/a. I'll do the second.

S = a + 64/a

Now, you can find the derivative of that equation, equate to zero and solve for a, the first number.

Then use the first equation to get the value of b.

Of course, you can get the answer by trying a few numbers, but the derivation is a sure way to get it :)

Can you post your results? :)

As for your second question... I don't understand what you mean by that :(

For the question posed in the body of the post: I believe the term "most economical dimensions" means using the least possible amount of material for the can. In other words - find the minimum possible surface area for a can that has volume = 1 quart. The area of the can is equal to the area of the bottom + lid + sides:


A = 2 \pi R^2 + 2 \pi R H


The volume must equal 1 quart, which is 57.75 cubic inches:


V = \pi R^2 H = 57.75\ in^3


rearrange to get:


H = \frac {57.75 in^3} {\pi R^2}


You can now substitute this value for H into the first equation to get the area as a function of R alone. Then take the derivative of A with respect to R and set that to zero - this will let you determine a value for R that yields either a minimum or maximum value of A. You can tell which it is by considering the second derivative derivateive - if the second derivative is positive at this value of R then you have the minimum for A, and have found the radius that gives the minimal amount of surface area for a quart can.

Hope this helps.

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Hi guys your anser is... I don't know