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Math Help - Optimization problems

  1. #1
    Junior Member
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    Sep 2007
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    Optimization problems

    1.) For its beef stew. Betty Moore Company uses aluminum containers that have the form of right circular cylinders. Find the radius and height of a container if it has a cpacity of 216cm^3 and is constructed using the least amount of metal.

    2.) A wooden beam has a rectangular cross section of height h cm and width w cm. The streghth of S of the beam is directly proportional to its width and the square if its height. What are the dimensions of the cross section of the strongest beam that can be cut from a round log of diameter 24cm?
    hint: * S=kh^2w, where k is a constant of proportionality.
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  2. #2
    Junior Member Serena's Girl's Avatar
    Joined
    Jul 2008
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    Solution to problem 1

    Solution for problem # 1:

    You are given a cylinder with a fixed volume of 216 cubic cm. You are tasked to find the radius and height which will require the minimum amount of metal (i.e. smallest surface area).

    In order to minimize the surface area, you have to obtain the derivative of the surface area with respect to one of the other dimensions, and then equate that derivative to zero. For example, if A is the surface area and R is the radius, then:

    <br />
\frac {dA} {dR} = 0<br />

    And then solve for the values of R and height (let's represent height with the variable h). Since the volume V of a circular cylinder is

    <br />
V = \pi R^2 h<br />

    then the value of h is

    <br />
h = \frac {216} {\pi R^2}<br />

    The total surface area of a circular cylinder is the sum of its lateral area and the areas of its bases:

    <br />
A = 2 \pi Rh + 2 \pi R^2<br />

    Substitute h into the equation for A and simplify, to obtain:

    <br />
A = \frac {512} {R} + 2 \pi R^2<br />

    Differentiate with respect to R and equate to 0:

    <br />
\frac {dA} {dR} = \frac {-512} {R^2} + 4 \pi R = 0<br />

    Solve for R, then solve for h.
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