MATH 112 - Winter 2005
Exam 2, Version 2 - Hints and Answers
1. (4 points each)
(a) f0(t) = (t4−6t3+ 4t)(5t4+ 28t3+ 4t)+(t5+ 7t4+ 2t2)(4t3−18t2+ 4)
(b) R0(q) = (1 + 3eq)·1
q2−5q
·(2q−5) −ln(q2−5q)·(3eq)
(1 + 3eq)2
(c) fx(x, y) = 2xey+ (ex2)(2x)y+ex2y(2xy)
(d) fy(x, y) = x2ey+ex2+ex2y(x2)
2. (10 points) HINT: The five vertices are (0,2), (0,10), (5,9), (7,0) and (3,0).
ANSWER: maximum=87; minimum=9
3. (4 points each)
(a) HINT: MC(q) = 0.03q2−0.6q+ 3. The question asks where this function has
a horizontal tangent line. So, you must compute the qat which MC0(q) = 0.
MC0(q) = 0.06q−0.6. Set this equal to 0 and solve for t.
ANSWER: q= 10
(b) i. ANSWER: global maximum= M C(14) = 0.48; global minimum= M C (11) =
0.03
ii. ANSWER: global maximum= M C(6) = 0.48; global minimum= M C (10) =
0
(c) ANSWER: P(q) = −0.01q3+ 0.22q2−0.65q−4
P0(q) = −0.03q2+ 0.44q−0.65. We need to know if Phas a horizontal tangent
line at q= 13. Plugging 13 into P0gives P0(13) = 0. So, Phas a horizontal
tangent line at q= 13.
P00(q) = −0.06q+ 0.44 Plugging 13 into P00 gives P00(13) = −0.34, which is
negative. So, Pis concave down at 13.
Since Phas a horizontal tangent and is concave down at q= 13, q= 13 gives a
local maximum of profit.
4. (4 points each)
(a) HINT: fx(x, y ) = x−5 + 2yand fy(x, y) = −y−4 + 2x. Set these partial
derivatives equal to 0 and solve the resulting system of equations.
ANSWER: (x, y) = (2.6,1.2)
(b) HINT: A≈fy(4,4) = 0; B≈fx(4,4) = 7
ANSWER: Bis bigger
(c) HINT: h0(1) = fy(6,1) = 7, k0(5) = fx(5,2) = 4
ANSWER: i is steeper