A rectangular beam is cut from a cylindrical log of radius 25 cm. A cross-section of the log is shown below. The strength, S, of a beam of width w and height h is proportional to wh2.
(a) Express the strength of the beam as a function of w only. Use k as the proportionality constant.
(b) Find the exact width and height of the beam of maximum strength.
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In typical 'optimization' problems, you want to
--- Choose the value of one variable, so that
--- some other variable is 'optimized'.
So you express the variable to be optimized in terms of the other variable. Sometimes there is more than one other variable. Then you need some fact that lets you eliminate all but one. So finally, you write something like:
V = some function of (r)
meaning you want to find a value of 'r' that maximizes (minimizes) V.
Once you get this far, the rest is downhill. (maybe a bumpy ride downhill, but you've done the hard part)
You differentiate, to get dV/dr.
You set the derivative = 0.
You solve for r.
You take your value(s) of r (maybe more than one) and substitute back into the function to get your optimum value. You ALSO check back with any endpoints for the practical situation. One of those is the answer.
In your example:
A college is planning to construct a new parking lot. The parking lot must be rectangular and enclose 6,000 sq. meters of land. A fence will surround the parking lot and another fence parallel to one of the sides will divide the parking lot into two sections. What are the dimensions, in meters, of the rectangular lot that will use the least amount of fencing.
You will draw an ACCURATE AND CLEAR diagram. You will find that, if we designate the width as W, and the height as H, there is an extra fence whose length is H.
<-------------- W -------------->
| | | ^
| | | |
| | | H
| | | |
| | | V
You want to minimize the total fencing:
F = 2W + 3H
Oops, that is expressing F in terms of two variables. Therefore, we have to use some fact in the example to eliminate a variable. OK, the area is 6000. Area of a rectangle is WH, so:
WH = 6000, or H = 6000/W
Put that in and we have:
F = 2W + 3(6000/W) = 2W + 18000W^-1 (best way to write it)
Now the problem is done. Well, there is a bit of grunt work coming, but we really have it solved. All we have to do is:
dF/dW = 2 - 18000W^-2 = 2 - -----
Set that equal to zero and solve:
2 - ----- = 0 --> w^2 = 9000 --> w = 30 sqrt(10)
And h = 6000/W = 200/sqrt(10) = 20 sqrt(10) -- remember rationalizing a denominator?