MATH 112 - Spring 2005
Exam II - Version 1
Hints and Answers
1. (4 points each)
(a) dy
dx =1
x1/3+e4x5·1
3x−2/3+e4x5
·20x4
(b) G0(t) = 10 "x3
1 + e−8x#9
·"(1 + e−8x)(3x2)−x3(e−8x)(−8)
(1 + e−8x)2#
(c) ∂z
∂y =−1
2hx2+ (4y+ 3)2i−3/2
·2(4y+ 3)(4)
(d) fx(x, y) = y[x2·ex+ex·2x]
2. (a) (4 points) Draw the vertical line x= 20, the horizontal line y= 8, and the line 5x+10y=
133 (which has intercepts (0,13.3) and (26.6,0)). The feasible region is a five-sided
polygon.
(b) (5 points) The feasible region has five vertices: (0,0), (0,8), (20,0), (10.6,8) and (20,3.3).
(c) (3 points) The optimum values of the objective function must occur at a vertex of the
feasible region. Plug all of your vertices into the objective function and choose the
largest and smallest values.
ANSWER: minimum = 0; maximum = 3.06
3. (a) (5 points) HINT: Find the profit function: P(q) = −0.1q3−2.4q2+ 86.4q−330.Take
the derivative: P0(q) = −0.3q2−4.8q+ 86.4. Set P0(q) = 0 and solve for q, using the
quadratic formula. You get one positive solution: q= 10.76. To apply the Second
Derivative Test, take P00(q): P00(q) = −0.6q−4.8. Since P00(10.76) is negative, profit
has a local maximum at q= 10.76.
ANSWER: q= 10.76 Items
(b) (5 points) HINT: AV C(q) = 0.1q2−0.6q+ 3.6. AV C 0(q) = 0.2q−0.6. Set AV C 0(q) = 0
and solve for q:q= 3. Compute AV C(1), AV C (3), and AV C(10).
ANSWER: smallest = $2.70 per Item; largest = $7.60 per Item
(c) (4 points) HINT: Marginal cost is the derivative of total cost: M C (q) = 0.3q2−1.2q+3.6.
Find MC0(q) and M C00(q): M C0(q) = 0.6q−1.2 and M C 00(q) = 0.6. So, MC 00(1) = 0.6.
Since MC00 (1) is positive, MC is concave up at q= 1.
ANSWER: concave up
4. (a) (4 points) HINT: Set the partial derivatives equal to 0 and solve the resulting system of
equations.
ANSWER: y= 1.525x+ 1.5625
(b) (2 points) HINT: E(1,2.003) −E(1,2)
0.003 ≈Em(1,2)
ANSWER: 59
(c) (2 points) HINT: You need ∂E
∂b to be positive. Choose any pair of numbers mand bthat
make 2b+ 15m−26 >0.
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