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Math 112 Solutions for 3.5, p. 269

3.
(a) about $125. (b) about 5.5 tons. (c) about 5.6 tons. (d) about 10 tons. (e) about 10.5 tons.
5.
The demand function is $ q=1280-160p$ and the revenue function is $ R=pq=1280p-160p^{2}$, whose maximum occurs at $ p=\$4.00$; the demand at that price is 640.
7.
(a) $ q\approx 555.56$. (b) $ q\approx 1414.214$. (c) $ q=1600$.
10.
(a) $ C(q)=49+.18q$, $ a(q)=\frac{49}{q}+.18$. (b) $ q=480-80p$. (c) $ R=480p-80p^{2}$. (d) $ p=\$3.00$ (e) $ p=\$3.09$.
12.
Let $ n$ be the number of times to order per year. Then the lot size is $ \frac{360}{n}$, the number on hand averages $ \frac{180}{n}$, and the inventory cost is $ C=\frac{180}{n}+5n+180$ whose minimum is at $ n=6$. Thus orders should be placed 6 times per year, with lot size 60.
14.
Line $ t_{1}$ has equation $ y=a(q_{0})+a'(q_{0})(q-q_{0})$, so has intercept $ b=a(q_{0})-q_{0}a'(q_{0})$. Since line $ t_{1}$ has slope $ a'(q_{0})$, line $ t_{2}$ has slope $ 2a'(q_{0})$ and equation $ y=2a'(q_{0})q+b= 2a'(q_{0})q+a(q_{0})-q_{0}a'(q_{0})$. Hence on $ t_{2}$ when $ q=q_{0},
y=2a'(q_{0})q_{0}+a(q_{0})-q_{0}a'(q_{0})=a(q_{0})+q_{0}a'(q_{0}).$ Now $ a(q)=\dfrac{C(q)}{q}$, so $ a'(q)=\dfrac{C'(q)q-C(q)}{q^{2}}$, so $ a'(q_{0})=\dfrac{C'(q_{0})q_{0}-C(q_{0})}{q_{0}^{2}}$. Hence $ y=a(q_{0})+q_{0}\cdot
\dfrac{C'(q_{0})q_{0}-C(q_{0})}{q_{0}^{2}}=
a(q_{0})+C'(q_{0})-\dfrac{C(q_{0})}{q_{0}}=C'(q_{0}).$

15.
(a) $ E=-\frac{-3}{47}\approx -.0638$.
17.
$ E=-\frac{5}{3} \approx -1.6667$.





Jason Grout 2003-03-03