5.2#15
a. If the graph shows the function: the inflection points are where you expect, basically - anywhere the graph goes from concave up to concave down, or vice versa. This graph has three.
b. If the graph shows the derivative f': the inflection points of f itself are where the derivative goes from increasing to decreasing or vice versa, i.e. local maxima and minima. This graph has two maxima and two minima.
c. If the graph shows f'': the inflection points of f itself are where f''=0, and is actually passing through 0 from positive to negative or vice versa. This graph has three zeroes, but only the first two are actually from passing through. The third is the sort of tricky behavior you'd get with e.g. x^4. It's as flat as your usual f''=0 inflection point, but it actually goes from concave up (f''>0) to flat to concave up again, and so isn't an inflection point.
#16.
y=x^3 is a nice such function (forgive me for not graphing it; I hope you know what it looks like).
#20. Again, I'll just describe these rather than actually graphing, since it's a pain to put graphs here. In both problems we can think about the volume of water after time t, call it V(t), and write it as some function of the depth.
a. The depth is linear in the time, d = kt, where k is some constant. (To be precise: it's proportional to the flow strength divided by the area of the base of the cylinder.)
b. V(t) = Ft, where F is the flow strength in e.g. gallons/minute. Then we have to figure out what the volume of a height d cone is. If one doubles the height, one also doubles the width and depth, so the volume goes up by 2^3. More generally V(t) = v d^3 for some v (the volume of a height-one cone, itself not so hard to figure out, but whatever). So d is proportional to the cube root of t. Since you know the graph of y=x^3 - it starts out flat and curves up, etc. - just flip it about the y=x axis to get this graph, which will therefore start with infinite derivative and curve down.
#26.
5.3#4,6.
#20.
We're making 75 cents profit by making this 2000th unit. So presumably if we made some more, we'd continue to increase our profit. Therefore, maximum profit probably comes above 2000.
(Why only "probably"? Maybe at 1900 we had to rent a new incredibly expensive warehouse, and profits went way down, and maybe we'll never make up for that cost. So assuming it's being rented, yeah, we should increase profits to a local maximum. But the global maximum was at 1900, when it wasn't being rented. Slightly tricky question.)
#23. Demand is p = b1 - a1 q, cost is C = b2 + a2 q. What q maximizes profit?
Profit = revenues - costs = pq-C = (b1 - a1 q)q - (b2 + a2 q) = (b1-b2)q - a1 q^2 - b2. So as with most of these book questions, we're finding the maximum point of a parabola, and from here it's plug and chug. The derivative of profit with respect to q is b1 - b2 - 2 a1 q, so is maximized at (b1-b2)/(2 a1). Let's not bother with the second derivative since it is after all a parabola. Oh, all right; the second derivative is -2 a1 which is negative so yes that was a local maximum.
5.5#10.
a. 5000-10(2^2) = 4960 pounds of yams.
b. E = -p/q dq/dp = -p/q (-20p). Plugging in p=2, q=4960, that's -2/4960 (-40) = 80/4960 = 1/62. Very small. Being so much less than 1 means that it's more accurate to say "People want yams and will buy them no matter what the price."
#11.
a. Revenue = pq = 2*4960 = $9920.
b. R = p(5000-10p^2). dR/dp = p(-20p) + (5000-10p^2) by product rule, which = 5000-30p^2. So it's zero if 5000=30p^2, or p = the square root of 5000/30 (which is about 12.9). Is this a maximum? The second derivative is -60 p, so is indeed less than zero at p = sqrt(5000/30) (which is how we'll write "square root" on the computer).
c. The quantity sold at that price is 5000-10p^2 = 5000 - 10*5000/30 = 5000*(2/3) = 10000/3. If you prefer to say 3333 1/3 feel free but it might be more annoying to plug into other formulae.
d. What's E=elasticity at that value? (Better be 1!) It's -p/q dq/dp = -p/q (-20p) = 20 p^2/q = 20 * (5000/30) / 10000/3 = 1. Sure enough.
#16.
If q = k/p^r = k p^-r, then dq/dp = -r k p^(-r-1) = -r k/p^(r+1). So E = (-p/q) dq/dp = (-p /(k/p^r)) (-r k/p^(r+1)). The p's and k's all cancel leaving only r.
#17.
If E > 1, people buy lots more if there's a sale, so one should lower prices. Mainly this problem serves to show that E = the constant 2 is ridiculous because it leads one to lower one's prices to zero, where people are buying infinitely much. In the real world they'd stop at some point and E would drop below 1.