The Corporeal Tutorial: Problem #10

This problem incorporates many different calculus preliminaries, specifically Trig preliminaries. It is an optimization problem that I think you will like (or at least appreciate).

A drainage channel is to be made so that its cross section is a trapezoid with equally sloping sides. If the sides and bottom all have a length of 5ft, how should the angle be chosen to yield the greatest cross-sectional area of the channel?

Sweet, I'm going to add my calculations and you make sure I didn't fuck up. Thanks. (warning: my variables are slightly different)

Area of a trapezoid:
$A(x)=\frac{1}{2}(10+2x)h$

Things we know:
$h=\sqrt{25-x^{2}}$

$x=cos(\theta )5$

Here we go :-)

$A(x)=25(1+cos (\theta) )\sqrt{1-cos^{2}(\theta )}$

$A(x)=25(1+cos (\theta) )\sqrt{sin^{2}(\theta )}$

$A(x)=25sin(\theta )(1+cos (\theta) )$

$A'(x)=25\left [cos(\theta )+cos^{2}(\theta )-sin^{2}(\theta ) \right ]$

$0=\left [cos(\theta )+cos^{2}(\theta )-1+cos^{2}(\theta ) \right ]$

$0=\left [2cos(\theta )-1 ]\left [ cos(\theta )+1 \right ]$

Before we solve for theta I should probably tell you our interval :-)

$\left ( 0\leq \theta \leq \frac{\pi }{2} \right )$

and evaluating our equations the answer:

$\therefore \frac{\pi }{3}$

The Corporeal Tutorial

Tuesday, June 23, 2009

Problem #10

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