7.4#16
@C/@d = 40 dollars/day, @C/@m = .15 dollars/mile. Each of these is the incremental cost of extending one's rental.
#20
N=80 measured in workers, V=30 measured in $25K of equipment. So f = 5 * 80^.75 * 30^.25 tons, or about 313 tons. @f/@N = f * .75/N tons/worker or about 2.9 tons/worker. @f/@V = f * .25/V tons/$25K of equipment or about 2.6 tons/$25K of equipment. They each mean how much production would increase, if you increased the inputs a little bit.
#28
@f/@x = 2/y. @f/@y = -2x/y^2. @^2f/@x^2 = 0. @^2f/@y^2 = 4x/y^3. @^2f/@x@y = @^2f/@y@x = -2/y^2.
#35
@f/@x = -8xt. @f/@t = 3t^2 - 4x^2. @^2f/@x^2 = -8t. @^2f/@t^2 = 6t. @^2f/@x@t = @^2f/@t@x = -8x.
7.5#9
@f/@x = -2x -B. For that to be 0 at (-2,1), B = 4. @f/@y = -2y -C. For that to be 0 at (-2,1), C = -2. f(-2,1) = A - (2^2 +4(-2) + 1^2 + (-2)1) = A + 5. For that to be 15, A = 10.
#12
This is a great question, I think. In part (a) we're thinking like an individual businessman - how can I maximize my own profit? In part (b) our viewpoint is more global - if these businessmen are each maximizing their profits, how will they behave when so-and-so changes?
a) Revenue = p1*q1 + p2*q2, profit = P = that - 2q1^2 - 2q2^2 - 10. @P/@q1 = p1 - 4q1, @P/@q2 = p2 - 4q2. So to be at a critical point, we need q1 = p1/4, q2 = p2/4.
The profit there is p1^2/4 + p2^2/4 - 2(p1/4)^2 - 2(p2/4)^2 - 10 = 1/8 (p1^2 + p2^2) - 10.
b) @maxprofit/@p1 = 1/4 p1.
8.1#3
dy/dt = 4 t^3, so t dy/dt = 4 t^4 = 4y.
#13
Plug y into the diff-eq. y' = 2x, so 10 = 2y - xy' = 2(x^2+k) - x (2x) = 2k. So k=5.
8.2#1
a) This is really too hard to draw here. Ask your TA at the review section.
b) Looks like y=-x-1.
c) If y=-x-1, y'=-1, which is indeed x+y. Ta-da!
#4
a) If the slope y'=xy, then at (2,1) it's 2, at (0,2) it's 0, at (-1,1) it's -1, and at (2,-2) it's -4.
b) You know how to do this - negative slopes point NW-SE, zero is flat, positive slopes point SW-NE.