Chapter 4.6, Problem 43E

a.

To determine

At a cost of dollars, a business can produce goods. Sales proceeds of dollars and a dollar profit (all amounts in thousands).

Answer to Problem 43E

The Chain Rule and the Power Rule to distinguish

  $0$ Of $\delta$

  $\text{0,3,0}\text{.9,0}\text{.6}$

Explanation of Solution

Given: Using the Power Rule and the Chain Rule to differentiate,

  $\frac{dc}{dt}=\left(3x^2-12x+15\right)\frac{dx}{dt}$

Substitute $\text{\$}\backslash x=2,\backslash \backslash drac\left\{dx\right\}\{ddt=0.1\text{\$}$

  $\begin{array}{c}{\left.\frac{dc}{dt}\right|}_{x=2}=(3\cdot 4-24+15)\cdot 0.1\\ =0.3\end{array}$

The Chain Rule is applied

  $\frac{dr}{dt}=9\frac{dx}{dt}$

Substitute $\$\backslash x=2,\backslash \backslash \$dfrac\$\left\{\text{dx}\right\}\left\{\text{dt}\right\}=0.1\$$

  ${\left.\frac{dr}{dt}\right|}_{x=2}=9\cdot 0.1=0.9$

Given that $p\left(x\right)=r\left(x\right)-c\left(x\right),$ differentiating

  $\begin{array}{c}\frac{d}{dt}p(x(t))=\frac{d}{dt}r(x(t))-\frac{d}{dt}c(x(t))\\ \frac{dp}{dt}=\frac{dr}{dt}-\frac{dc}{dt}\end{array}$

Value of $\text{x}$

If $\text{x=2,}$ then using the values previously calculated

  ${\left.\frac{dp}{dt}\right|}_{x=2}=0.9-0.3=0.6$

Conclusion:

The Chain Rule and the Power Rule to distinguish

  $0$ Of $\delta$

  $\text{0,3,0}\text{.9,0}\text{.6}$

b.

To determine

At a cost of dollars, a business can produce goods. Sales proceeds of dollars and a dollar profit (all amounts in thousands).

Answer to Problem 43E

Using the previously obtained values, for $x=1.5$ ,

  $\frac{dp}{dt}=5.0625$

Explanation of Solution

Given: Using differentiation and the Chain and Power.

  $\begin{array}{c}\frac{dc}{dt}=\left(3x^2-12x-\frac{45}{x^2}\right)\\ \frac{dx}{dt}=\left(3{(1.5)}^2-12(1.5)-\frac{45}{{1.5}^2}\right)(0.05)\\ =-1.5625\\ \frac{dr}{dt}=70\\ \frac{dx}{dt}=70(0.05)\\ =3.5\\ \frac{dp}{dt}=\frac{dr}{dt}-\frac{dc}{dt}\\ =3.5-(-1.5625)\\ =5.0625\end{array}$

Conclusion:

Using the previously obtained values, for $x=1.5$ ,

  $\frac{dp}{dt}=5.0625$