Question

From Unit 2 Exercises: Applications of Differentiation D. More Max-min Problems
Questions 1 and 2 are bothering me for many days now.

http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-2-applications-of-differentiation/part-b-optimization-related-rates-and-newtons-method/problem-set-4/MIT18_01SC_pset2prb.pdf

Answer 1

For problem 2D-1
let f(a) be the lift/drag, i.e. a function of attack angle "a" that we wish to find the min and max.

in this case
\[ f(a) = a (a^2+\tau^2)^{-1} \]
where \( \tau \) is a constant.

Take the usual approach
\[ \frac{d}{da} f(a) = 0 \]
and solve for a

what do you get ?

Answer 2

for problem 2D-2
The volume of the cone is
\[ V(a) = \frac{1}{3} \pi r^2 \cdot a r = \frac{\pi }{3} a r^3 \]
Find the "a" that maximizes the volume.
the surface area A is constant, so we can solve for r in terms of a:
\[ A = \pi r^2 (1+a^2)^\frac{1}{2} \\
r^3 = \left(\frac{A}{\pi}\right)^\frac{3}{2}(1+a^2)^{-\frac{3}{4}}
\]

substitute for r^3 in the formula for the volume of the cone, to get a function that is only a function of "a". Take the derivative dV/da (i.e. with respect to "a"), set equal to zero, and solve for a.

Answer 3

Thanks for the reply.

Answer 4

What I got in D-1 was the best attack angle would be \[\tau\] and for the minimum ratio, it is \[\frac{ 1 }{ 2\tau }\]. But what would this angle look like?

Answer 5

Yes, the max lift occurs at + tau (and the minimum "lift" at - tau)

tau is the angle between the side and the diagonal (as you show in your diagram)
attack angle "a" is the angle between the horizontal and the diagonal