MATH 112 – EXAM II Hints and Answers
Version Alpha
Winter 2006
1. (5 points each)
(a) dx
dy =x3·12(4x+ 9)11(4) + (4x+ 9)12 ·3x2
(b) f0(z) = 6[ln(7ez+z2)]5·1
7ez+z2·(7ez+ 2z)
(c) dy
dx =√6x+ 1[e3x·4x3+ (x4+ 5)e3x·3] −e3x(x4+ 5) ·1
2(6x+ 1)−1/26
(√6x+ 1)2
2. (a) (2 points) HINT: h(q) is a quadratic function whose graph is a parabola that opens
upward. Find its vertex.
ANSWER: from q= 0 to q= 60 Things
(b) (2 points) ANSWER: T R(q) = h(q)·q=q3−120q2+ 3600q
(c) (6 points) HINT: Compute T R0(q), set it equal to 0, and solve for qto find the critical
numbers of T R. Plug the critical numbers and the endpoints of the interval back into
the formula for T R.
ANSWER: q= 20 Things
(d) (2 points) ANSWER: h(20) = 1600 dollars per Thing
3. (a) (4 points) ANSWERS: fx(x, y) = 10x−39.8 + y;fy(x, y) = 12y−73 + x
(b) (5 points) HINT: Set fx(x, y) = 0 and fy(x, y) = 0 and solve the resulting system of
equations.
ANSWER: (x, y) = (3.4,5.8)
(c) (4 points) HINT: The steepness of the graph of g(t) at t= 10 is measured by the slope
of the tangent line to g(t) at t= 10. That’s g0(10), which is equal to fx(10,15) = 75.2.
Similarly, the steepness of the graph of h(t) at t= 15 is measured by h0(15), which is
fy(10,15) = 117.
ANSWER: B
4. (10 points) HINTS: The constraints are x≤200, y≤150, and x+y≤280. The objective
function is P(x, y) = 1.25x+ 1.40y. The feasible region is a five-sided polygon with vertices
(0,0), (0,150), (200,0), (200,80), and (130,150).
ANSWER: x= 130 and y= 150 bags