Homework 7 - Solution

Date: MATH 1100-2 - Spring 2002

1. 10.3.1. We start by looking for the critical values. $f(x) = x^3-2x^2-4x+2$, so that $f'(x) = 3x^2-4x-4$. To solve $f'(x)=0$ we use the quadratic formula and find that the critical values are $x=-\frac{2}{3}$ and $x=2$.

   To find the absolute maximum and minimum we compare the values of $f$ at the critical values and the values at the endpoints of the interval. The highest value is the absolute maximum and the lowest the absolute minimum. From Table 1 we conclude that the maximum happens for $x=-\frac{2}{3}$ and the minimum is $x=2$.
   

   Table 1: Values of $f$ for exercise 10.3.1.

   $x$	$-1$	$-\frac{2}{3}$	$2$	$3$
   $f(x)$	$3$	$\frac{94}{27}\sim 3.48$	$-6$	$-1$

   
   

2. 10.3.3. We proceed as in the previous exercise. $f(x)=x^3+x^2-x+1$, so that $f'(x)=3x^2+2x-1$ and the critical values are $x=-1$ and $x=\frac{1}{3}$. The second of these values can be discarded because it is not in the interval under consideration. Next we produce a table of values of $f$ at the critical values and at the endpoints of the interval. Using Table 2 we conclude that the maximum is at $x=-1$ and the minimum at $x=-2$.
   

   Table 2: Values of $f$ for exercise 10.3.3.

   $x$	$-2$	$-1$	0
   $f(x)$	$-1$	$2$	$1$

   
   

3. 10.3.5.
   1. Maximizing revenue means maximizing $R = 36x-0.01x^2$ (for $x\geq 0$). Since $R' = 36-0.02 x$, we see that the only critical value of $R$ is $x=1800$. To see if that value is a local maximum or minimum we use the second derivative test: since $R'' = -0.02<0$, we conclude that $x=1800$ is a local maximum. Actually, from the fact that the graph of $R$ is a parabola pointing downwards we immediately conclude that the local minimum is actually the absolute minimum. Thus, the minimum revenue is $R(1800) = 32400$.
   2. If the production is limited to $x\in [0,1500]$, the absolute maximum computed above cannot be reached (it is not in the region under consideration). So we have to consider the values of $R$ at the endpoints $x=0$ and $x=1500$ (there are no critical values between them). Since $R(0)=0$ and $R(1500)=31500$, we conclude that the maximum revenue for a production of up to $x=1500$ is $31500$.

4. 10.3.25. Since profit $P(x)= R(x) - C(x) = 4600x-(45000+100x+x^3) = -x^3+4500x-45000$, we look for the critical values of $P$ by solving $P'(x) = 0$. Since $P'(x) = -3x^2+4500$, the critical values occur at $x=\pm \sqrt{1500} \simeq \pm 38.73$. $x$ being a number of units, it has to be positive, so the only interesting critical value happens at $x\simeq 38.73$. The second derivative test says that since $P''(38.73) = -0.02 < 0$, the point is a local maximum.

   But, $x=38.73$ is not an integer! So we check the two closest integers to see which one has the higher value: $P(38)=1353.56$ and $P(39)= 1388.79$. Then we conclude that the maximum profit is $1388.79$, which happens when the production level is $x=39$.

5. 10.4.1.
   1. For each district we have to find first the critical values of the sales function. Since $S_1 = 30+20x_1-0.4x_1^2$ and $S_2=20+36x_2-1.3x_2^2$, we have $S'_1 = 20-0.8x_1$ and $S'_2 = 36-2.6x_2$, we conclude that the critical values are $x_1=25$ and $x_2 = 13.8462$. Since we have $S_1'' = -0.8<0$, and $S_2''=-2.6$, we conclude (using the second derivative test) that both critical values are local maxima. Therefore, the sales revenue are maximized by spending $x_1=25$ and $x_2 = 13.8462$ (millions of dollars in each case).
   2. This is just the sum of the amounts needed to maximize each region: $38.8462 = 25+13.8462$ millions of dollars.

6. 10.4.5.
   1. Since the hourly number of units produced after $t$ hours is given by $y(t) = 70t+\frac{1}{2}t^2-t^3$, what we want to do is to maximize this function on the interval $[0,8]$. We start by finding the critical values, that is, solving $R'(t)=0$. Since $R' = 70+t-3t^2$, we see that the critical points would be $t=-\frac{14}{3}$ and $t=5$. But, the first is not in the region under consideration. Therefore, the only critical point is $t=5$. From Table 3 we can compare the values of $y$ at the critical point and at the two endpoints. We conclude that the maximum of $y$ happens after $t=5$ hours.
      

      Table 3: Values of $y$ for exercise 10.4.5.

      $t$	0	$5$	$8$
      $y(t)$	0	$237.5$	$80$

      
      

   2. That is $y(5) = 237.5$, reading from Table 3.

7. 10.4.7. That is, we want to find the price $p$ that maximizes the expenditure function $E(p) = 10000p-100p^2$. We look for the critical points: $E'=10000-200p$, so that the critical value is $p=50$. Since $E''=-200<0$, by the second derivative test we conclude that $50$ is a local maximum. Actually, since the graph of $E$ is that of a parabola pointing down we conclude that $50$ the maximum of $E$ is achieved when the price is $p=50$.

8. 10.4.13. We have to find the maximum of the function $p(t) = \frac{6.4t}{t^2+64}+0.05$ (for $t\geq 0$). We start by looking for the critical values.

   $$p'(t) = \frac{6.4(t^2+64) - 6.4t \cdot 2t}{(t^2+64)^2} = \frac{- 6.4t^2+409.6}{(t^2+64)^2}.$$

   
   
   Then, the critical values satisfy: $-6.4t^2+409.6=0$, and they are $t= \pm 8$. Clearly, $t=-8$ is not a possibility in the context of this problem, and the only critical value is $t=8$. We can use the first derivative test to conclude that $t=8$ is a local maximum. Then, the maximum value of $p$ happens when $t=8$ and that value is $p(8) = 0.45$.

9. 10.4.31. The cost function that we want to minimize is composed of two parts: plate preparation cost ($2\cdot x$), and press use cost, $12.5\cdot h$, where $h$ is the number of press-hours needed to complete the job. Thus, in principle, the cost is

   $$C = 2x+ 12.5 h.$$

   
   

   There is a relation between $x$ and $h$: if $x$ plates are used and $1000$ press impressions can be made per hour, in order to produce $100000$ posters, $\frac{100000}{1000x}$ press hours are needed. That is:

   $$h = \frac{100000}{1000x}.$$

   
   

   All together, the cost function becomes:

   $$C = 2x + 12.5\cdot \frac{100}{x} = 2x + \frac{1250}{x}.$$

   
   

   Now, we want to find the critical values of $C$. $C'=2-\frac{1250}{x^2}$, so that $C'=0$ means that $2-\frac{1250}{x^2} = 0$, which says that $x = \pm \sqrt{625} = \pm 25$. Since $x$ cannot be negative, the only interesting critical value is $x=25$.

   We use the second derivative test to check if $x=25$ is a local minimum: $C'' = (2-\frac{1250}{x^2})' = \frac{2500}{x^3}$. Then, $C''(25) = 0.16 >0$,and $x=25$ is a local minimum, as we wanted.

10. 10.5.1.
    1. From the graph, there is a vertical asymptote at $x=2$. From the formula $f(x)=\frac{x-4}{x-2}$, we see that the denominator vanishes at $x=2$, whereas the numerator doesn't, thus producing a vertical asymptote when $x=2$.
    2. From the graph it looks like $\lim_{x\rightarrow +\infty} f(x)= 1$. From the formula we compute

       $$\lim_{x\rightarrow +\infty} f(x) = \lim_{x\rightarrow +\infty} \f... ...\rightarrow +\infty} \frac{1-\frac{4}{x}}{1-\frac{2}{x}} = \frac{1-0}{1-0} = 1.$$

       
       
       Notice that the previous computation also computes $\lim_{x\rightarrow -\infty} f(x)=1$.
    3. From the graph, it looks like there is a horizontal asymptote with $y=1$. From the formula, the previous item shows that this is, in fact, the case.
    4. From the graph it looks like $\lim_{x\rightarrow -\infty} f(x)=1$. As we noted above this is also the result of the computation using the formula.

11. 10.5.5. For $y=\frac{2x}{x-3}$, we see that the denominator vanishes when $x=3$. Since the numerator doesn't vanish at $x=3$, we conclude that $x=3$ is a vertical asymptote.

    To detect horizontal asymptotes we have to compute:

    $$\lim_{x\rightarrow\infty} \frac{2x}{x-3} = \lim_{x\rightarrow\inf... ...3}{x}} = \lim_{x\rightarrow\infty} \frac{2}{1-\frac{3}{x}} = \frac{2}{1-0} = 2.$$

    
    
    Then, there is a horizontal asymptote at $y=2$.

12. 10.5.9. For $y=\frac{3x^3-6}{x^2+4}$, we look for zeros of the denominator $x^2+4$. But this is always a positive quantity that can't vanish. Therefore, there are no vertical asymptotes.

    To detect horizontal asymptotes we have to compute:

    $$\lim_{x\rightarrow\infty}\frac{3x^3-6}{x^2+4} = \lim_{x\rightarro... ... = \lim_{x\rightarrow\infty} \frac{3x-\frac{6}{x^2}}{1+\frac{4}{x^2}} = \infty.$$

    
    
    Therefore, there is no horizontal asymptote.