Question

3.4 rates of change

1. suppose that you run a toy factory. the total cost of producing $x$ toys is $c(x)=50,000 + 2x$ dollars, and the total revenue generated by selling $x$ toys is $r(x)=10x-.0001x^{2}$ dollars.

a. find $r(x)$, and interpret its meaning.
b. find $c(x)$, and interpret its meaning.
c. find $\frac{d}{dx}(r(x)-c(x))$, and interpret its meaning.
d. when is the derivative you found in part c positive, and when is it negative?
e. if your company wants to maximize total profits, how many toys should they produce? what is the maximum total profit? explain your reasoning.

Explanation:

Step1: Find $R'(x)$

Differentiate $R(x)=10x - 0.0001x^{2}$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$.
$R'(x)=\frac{d}{dx}(10x)-\frac{d}{dx}(0.0001x^{2})=10-0.0002x$. It represents the marginal revenue, which is the additional revenue generated by selling one more toy.

Step2: Find $C'(x)$

Differentiate $C(x)=50000 + 2x$. Since $\frac{d}{dx}(a)=0$ for a constant $a$ and $\frac{d}{dx}(ax)=a$, $C'(x)=\frac{d}{dx}(50000)+\frac{d}{dx}(2x)=2$. It represents the marginal cost, which is the additional cost of producing one more toy.

Step3: Find $\frac{d}{dx}(R(x)-C(x))$

First, $R(x)-C(x)=(10x - 0.0001x^{2})-(50000 + 2x)=-0.0001x^{2}+8x - 50000$.
Then, $\frac{d}{dx}(R(x)-C(x))=\frac{d}{dx}(-0.0001x^{2}+8x - 50000)=-0.0002x + 8$. This represents the marginal profit, which is the additional profit from producing and selling one more toy.

Step4: Determine when $\frac{d}{dx}(R(x)-C(x))$ is positive and negative

Set $-0.0002x + 8>0$.
$-0.0002x>-8$, then $x < 40000$. So the marginal profit is positive when $x < 40000$.
Set $-0.0002x + 8<0$.
$-0.0002x<-8$, then $x>40000$. So the marginal profit is negative when $x > 40000$.

Step5: Maximize the profit function

The profit function is $P(x)=R(x)-C(x)=-0.0001x^{2}+8x - 50000$.
The function is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-0.0001$, $b = 8$, $c=-50000$.
The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate at $x=-\frac{b}{2a}$.
$x=-\frac{8}{2\times(-0.0001)} = 40000$.
Substitute $x = 40000$ into $P(x)$:
$P(40000)=-0.0001\times(40000)^{2}+8\times40000 - 50000=-0.0001\times1600000000+320000 - 50000=-160000+320000 - 50000 = 110000$.

Answer:

A. $R'(x)=10 - 0.0002x$, marginal revenue.
B. $C'(x)=2$, marginal cost.
C. $\frac{d}{dx}(R(x)-C(x))=-0.0002x + 8$, marginal profit.
D. Positive when $x < 40000$, negative when $x > 40000$.
E. Produce 40000 toys. The maximum total profit is $110000$ dollars. The profit function is a quadratic function, and its maximum occurs at the vertex of the parabola. The $x$ - value of the vertex of the quadratic profit function gives the number of toys to produce for maximum profit, and substituting this value into the profit function gives the maximum profit.