Homework 6 - Solution

Date: MATH 1100-2 - Spring 2002

1. 11.4.1. For $y=x^3-3x$, we start by taking $\frac{d}{dt}$ on both sides:

   $$\frac{dy}{dt} = \frac{d}{dt}(x^3-3x) = \frac{d(x^3)}{dt} -\frac{d(3x)}{dt} = 3x^2 \frac{dx}{dt} -3\frac{dx}{dt}.$$

   
   
   We now substitute $x=2$ and $\frac{dx}{dt} = 4$ and obtain $\frac{dy}{dt} = 3\cdot 2^2\cdot 4 - 3\cdot 4 = 48 - 12 = 36$.

2. 11.4.17. Both $P$ and $x$ are functions of time ($t$). We start by applying $\frac{d}{dt}$ to both sides of $P = 180x-\frac{x^2}{1000} - 2000$:

   $$\frac{dP}{dt} = \frac{d}{dt}(180x-\frac{x^2}{1000} - 2000) = 180 ... ...0}\frac{dx^2}{dt} - 0 = 180 \frac{dx}{dt} - \frac{1}{500} x\cdot \frac{dx}{dt}.$$

   
   
   Now we can replace $x = 100$ and, since $x$ is increasing at a rate of $10$ units per day, we have that $\frac{dx}{dt} = 10$.

   $$\frac{dP}{dt} = 180\cdot 10 - \frac{1}{500}\cdot 100\cdot 10 = 1798.$$

   
   
   Thus, the profit is increasing at $1798$ dollars per day.

3. 11.4.19. We want to know the rate at which the price is changing. That is, we want to know $\frac{dp}{dt}$. Start by applying $\frac{d}{dt}$ to both sides of $p=\frac{1000-10x}{400-x}$:

   $$\begin{split}\frac{dp}{dt} &= \frac{d}{dt} \frac{1000-10x}{40... ...{(400-x)^2} = \frac{-3000\frac{dx}{dt}}{(400-x)^2}. \end{split}$$

   
   
   Now, plugging in $x=20$ and $\frac{dx}{dt} = -20$ (notice the negative sign because the demand is decreasing), we have

   $$\frac{dp}{dt} = -\frac{3000\cdot (-20)}{(400-20)^2} = \frac{150}{361} \simeq 0.4155.$$

   
   
   Thus, the price is increasing at approximately $42$ cents per day.

4. 10.1.3.
   1. $f'$ changing from positive to negative means that $f$ changes from increasing to decreasing, which happens at $x=1$.
   2. $f'$ changing from negative to positive means that $f$ changes from decreasing to increasing, which happens at $x=4$.
   3. $f'$ doesn't change sign at $x=-1$, $x=0$ and $x=5$ since near these points the function is increasing.

5. 10.1.5.
   1. The critical values of $f$ are the points where the derivative vanishes: $x=3$ and $x=7$.
   2. $f$ increases where $f'(x)>0$, that is, on $(3,7)$.
   3. $f$ decreases where $f'(x)<0$, that is, on $(-\infty,3)$ and on $(7,\infty)$.
   4. At $x=7$ $f$ goes from increasing to decreasing, so that $x=7$ is a relative maximum.
   5. At $x=3$ $f$ goes from decreasing to increasing, so that $x=3$ is a relative minimum.

6. 10.1.7. The critical values of $f(x)=2x^3-12x^2+6$ are those values of $x$ where $f'(x)=0$ or $f'(x)$ is not defined. Being $f$ a polynomial, $f'(x) = 6x^2-24x$ is always defined. So, the critical values arise as solutions of $f'(x)=0$:
   

   $$6x^2-24x =$$ 0

   $$x^2-4x =$$ 0

   $$x(x-4) =$$ 0

   
   
   So that the critical values are $x=0$ and $x=4$.

7. 10.1.9. In the previous exercise we found the critical values of $f$. Now we find the signs of the derivative in the corresponding intervals. We have:
   

   $$f'(-1) = 6+24=30 >0$$

   $$f'(1) = 6-24 = -18<0$$

   $$f'(5) = 30 >0$$

   
   

   Figure 1: Sign diagram for exercise 10.1.9

   
   The sign diagram is shown in Figure 1.

8. 10.1.17. For $y=\frac{x^3}{3} +\frac{x^2}{2} -2x+1$ we have
   1. $y' = x^2+x-2$.
   2. Since the derivative is always defined, we have to solve $f'(x)=0$, that is, $x^2+x-2=0$, which, for instance using the quadratic formula, has solutions $x=-2$ and $x=1$. Thus, the critical values are $x=-2$ and $x=1$.
   3. Evaluating the critical values we obtain the critical points: $(-2,\frac{13}{3})$ and $(1,-\frac{1}{6})$.
   4. Since we know the critical points, we evaluate $f'$ at some intermediate points:
      

      $$f'(-3) = 4>0$$

      $$f'(0) = -2 <0$$

      $$f'(2) = 4>0$$

      
      
      Thus $f$ is increasing on $(-\infty,-2)$ and $(1,\infty)$, while it is decreasing on $(-2,1)$.
   5. Since $f$ goes from increasing to decreasing at $x=-2$, we conclude that $x=2$ is a relative maximum. Conversely, since $f$ goes from decreasing to increasing at $x=1$ we conclude that $x=1$ is a local minimum. The graph of $f$ can be seen on Figure 2.

      Figure 2: Graph of $f$ for exercise 10.1.17 (e)

      

9. 10.1.29. We start by computing the derivative of $f(x)=3x^5-5x^3+1$: $f'(x) = 15x^4-15x^2$.

   Since the derivative is defined always (can be computed always), the critical values are the solutions of $f'(x)=0$. That is, $15x^4-15x^2 =0$, or $x^4-x^2=0$. To solve this last equation we take $x^2$ as common factor: $x^2(x^2-1)=0$. Now, if a product is 0, at least one of the two terms must vanish, so that $x=0$ or $x^2-1=0$, which has solutions $x=1$ and $x=-1$. Thus, the critical values are $x=-1$, $x=0$ and $x=1$.

   Now that we have the critical values, we can find where the function is increasing and decreasing by evaluating the derivative at some intermediate points:
   

   $$f'(-2) = 180>0$$

   $$f'(-\frac{1}{2}) = -\frac{45}{16}<0$$

   $$f'(\frac{1}{2}) = -\frac{45}{16}<0$$

   $$f'(2) = 180>0$$

   
   
   We conclude that $f$ is increasing on $(-\infty,-1)$ and on $(1,\infty)$, while it is decreasing on $(-1,1)$.

   With the previous data we see that $f$ goes from increasing to decreasing at $x=-1$, so that it is a local maximum; $f$ is decreasing on both left and right of $x=0$, so that $x=0$ is an inflection point and since $f$ goes from decreasing to increasing at $x=1$, that point is a local minimum. These conclusions are reflected by the graph shown in Figure 3.

   Figure 3: Graph of $f$ for exercise 10.1.29

   

10. 10.2.5. $f(x)$ is concave down to the left of $c$, and on $(d,e)$.

11. 10.2.7. $f''(x)>0$ means that $f$ is concave up, which happens on $(c,d)$ and to the right of $e$.

12. 10.2.9. These are points where the concavity changes: $c$, $d$ and $e$.

13. 10.2.19. We start by finding the first derivative of $y=x^4-16x^2$:

    $$\frac{dy}{dx} = 4x^3-32x.$$

    
    
    Then, the critical values are the zeroes of $y'$, that is $4x^3-32x = 0$, or $4x(x^2-8)=0$. Then, the critical values are $x=0$ and $x=\pm\sqrt{8}$. We want to use the second derivative test to decide what kind of points are these.

    $$\frac{d^2y}{dx^2} = \frac{d}{dx}(4x^3-32x) = 12x^2-32.$$

    
    
    Then,
    

    $$y''(-\sqrt{8}) = 64> 0$$

    $$y''(0) = -32 <0$$

    $$y''(\sqrt{8}) = 64> 0$$

    
    
    Thus, $x=0$ is a local maximum and $x=\pm\sqrt{8}$ are local minima. Our conclusions are shown in Figure 4.

    Figure 4: Graph of $f$ for exercise 10.2.19

    

14. 10.2.37.
    1. Production is maximized when we find a maximum of $P(t) = 27t+12t^2-t^3$. We start by finding $\frac{dP}{dt} = 27+24t-3t^2$.

       To find the maximum, we start by looking for critical values: $\frac{dP}{dt} =0$ so that $27+24t-3t^2=0$. Using the quadratic formula (or any other method), we find that $t = -1$ and $t=9$. Clearly, none of the solutions is acceptable. But then, if we evaluate the derivative at any point between $-1$ and $9$ (for instance $t=0$) we see that $\frac{dP}{dt} >0$, so that the production is increasing all the time: the maximum production will be at the end of the $8$ hour shift (which makes intuitive sense, unless the worker destroys part of his work during the day!).

    2. To maximize the rate of production, we want to maximize the function $R(t) = \frac{dP}{dt} = 27+24t-3t^2$. We start by computing its derivative: $R'(t) = 24-6t$. The critical points are the solutions of $24-6t=0$, that is $t=4$. Since $R''(t) = -6<0$, the second derivative test says that $t=4$ is a local maximum of $R$, that is, of the production rate.