Chapter 4.6, Problem 44E

a.

To determine

To calculate:Show that if the profit $P\left(x\right)$ is a maximum, then the marginal revenue equal to the marginal cost.

Answer to Problem 44E

The marginal revenue equal to the marginal cost i.e, $R^{\prime }\left(x\right)=C^{\prime }\left(x\right)$

Explanation of Solution

Given information:

The profit $P\left(x\right)$ is a maximum.

Formula used:

Total profit = marginal revenue − marginal cost

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Recall that,

Total profit = marginal revenue − marginal cost

  $P\left(x\right)=R\left(x\right)-C\left(x\right)$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate on both sides,

  $P^{\prime }\left(x\right)=R^{\prime }\left(x\right)-C^{\prime }\left(x\right)$

Solve for $P^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}{R}^{\prime }\left(x\right)-C^{\prime }\left(x\right)=0\\ \Rightarrow R^{\prime }\left(x\right)=C^{\prime }\left(x\right)\end{array}$

Conclusion:

Thus the marginal revenue equal to the marginal cost i.e, $R^{\prime }\left(x\right)=C^{\prime }\left(x\right)$

b..

To determine

If $C\left(x\right)=16,000+500x-1.6x^2+0.004x^3$ is the cost function and $p\left(x\right)=1700-3x$ is the demand function, find the production level that will maximize profit.

Answer to Problem 44E

The production level that will maximize profitis $x=100$ .

Explanation of Solution

Given information:

  $C\left(x\right)=16,000+500x-1.6x^2+0.004x^3$ is the cost function and $p\left(x\right)=1700-3x$ is the demand function.

Formula Used:

Revenue function = The product of per unit times the number of unit sold

Quadratic formula: let the quadratic equation is $a{x}^2+bx+c=0,\left(a\ne 0\right)$ Thus, if $b^2-4ac\ge 0$ then the roots of the quadratic equation is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

  $C\left(x\right)=16,000+500x-1.6x^2+0.004x^3$

The marginal cost is

  $C^{\prime }\left(x\right)=500-3.2x+0.012x$

Recall that,

Revenue function = The product of per unit times the number of unit sold

  $\begin{array}{l}R\left(x\right)=x\times p\left(x\right)\\ =x\left(1700-7x\right)\\ =1700x-7x^2\end{array}$

Marginal revenue is

  $R^{\prime }\left(x\right)=1700-14x$

Recall part a,

The marginal revenue equal to the marginal cost $R^{\prime }\left(x\right)=C^{\prime }\left(x\right)$

  $\begin{array}{l}500-3.2x+0.012x^2=1700-14x\\ \Rightarrow 0.012x^2+10.8x-1200=0\end{array}$

Recall that,

Quadratic formula: let the quadratic equation is $a{x}^2+bx+c=0,\left(a\ne 0\right)$ Thus, if $b^2-4ac\ge 0$ then the roots of the quadratic equation is given by

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $\begin{array}{l}0.012x^2+10.8x-1200=0\\ x=\frac{-10.8\pm \sqrt{{\left(10.8\right)}^2-4\times 0.012\times -1200}}{2\times 0.012}\\ =\frac{-10.8\pm \sqrt{174.24}}{0.024}\end{array}$

Solve for different sign’s

  $\begin{array}{l}0.012x^2+10.8x-1200=0\\ x=\frac{-10.8\pm \sqrt{{\left(10.8\right)}^2-4\times 0.012\times -1200}}{2\times 0.012}\\ =\frac{-10.8\pm \sqrt{174.24}}{0.024}\\ x=\frac{-10.8+\sqrt{174.24}}{0.024}\text{or }x=\frac{-10.8-\sqrt{174.24}}{0.024}\\ x=100\text{or }x=-1000\end{array}$

Take positive number

Conclusion:

Thus the production level that will maximize profit is $x=100$