28 points total
1) the cost to manufacture badminton rackets at a company in japan can be modelled by
the cost function ( c(x)=8 x^{3}-24 x^{2}+28 x ) where ( x ) is in hundreds of rackets and ( c(x) ) is
in hundreds of japanese yen (jpy).
a) find ( c^{prime}(x) ).
2
the marginal cost of production is the cost of producing one additional unit. this can be
approximated by the gradient of the cost function.
b) find the marginal cost when 100 rackets are produced and interpret its meaning
in this context.
3
the revenue from selling the rackets is given by the function ( r(x)=26 x ) where ( x ) is in
hundreds of rackets and ( r(x) ) is in hundreds of jpy.
c) given that profit = revenue - cost, determine a function for the profit ( p(x) ), in
hundreds of jpy from selling ( x ) hundreds of badminton rackets.
2
d) find ( p^{prime}(x) ).
2
e) determine the intervals where ( p(x) ) is increasing and decreasing.
4
the derivative ( p^{prime}(x) ) gives the marginal profit. the production will reach its optimal level
when the marginal profit is zero and ( p(x) ) is positive.
f) find the optimal production level and the expected profit at this level.
3

Question

Accepted Answer

### a) # Explanation: ## Step1: Differentiate term - by - term Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, for $C(x)=8x^{3}-24x^{2}+28x$: $C^{\prime}(x)=\frac{d}{dx}(8x^{3})-\frac{d}{dx}(24x^{2})+\frac{d}{dx}(28x)$ $C^{\prime}(x)=3\times8x^{2}-2\times24x + 28$ $C^{\prime}(x)=24x^{2}-48x + 28$ # Answer: $C^{\prime}(x)=24x^{2}-48x + 28$ ### b) # Explanation: ## Step1: Substitute $x = 1$ into $C^{\prime}(x)$ Since $x$ is in hundreds of rackets and $100$ rackets mean $x = 1$. $C^{\prime}(1)=24\times(1)^{2}-48\times(1)+28$ $C^{\prime}(1)=24 - 48+28$ $C^{\prime}(1)=4$ ## Step2: Interpret the result The marginal cost when $100$ rackets are produced is $4\times100 = 400$ JPY. It means that when the production level is $100$ rackets, the cost of producing one more racket is approximately $400$ JPY. # Answer: The marginal cost is $400$ JPY. It represents the approximate cost of producing one additional racket when $100$ rackets have already been produced. ### c) # Explanation: ## Step1: Use the profit formula $P(x)=R(x)-C(x)$ Given $R(x)=26x$ and $C(x)=8x^{3}-24x^{2}+28x$ $P(x)=26x-(8x^{3}-24x^{2}+28x)$ $P(x)=26x - 8x^{3}+24x^{2}-28x$ $P(x)=-8x^{3}+24x^{2}-2x$ # Answer: $P(x)=-8x^{3}+24x^{2}-2x$ ### d) # Explanation: ## Step1: Differentiate $P(x)$ term - by - term Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ for $P(x)=-8x^{3}+24x^{2}-2x$ $P^{\prime}(x)=\frac{d}{dx}(-8x^{3})+\frac{d}{dx}(24x^{2})-\frac{d}{dx}(2x)$ $P^{\prime}(x)=-3\times8x^{2}+2\times24x-2$ $P^{\prime}(x)=-24x^{2}+48x - 2$ # Answer: $P^{\prime}(x)=-24x^{2}+48x - 2$ ### e) # Explanation: ## Step1: Find the critical points Set $P^{\prime}(x)=0$, so $-24x^{2}+48x - 2 = 0$. Divide through by $-2$ to get $12x^{2}-24x + 1=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 12$, $b=-24$, $c = 1$ $x=\frac{24\pm\sqrt{(-24)^{2}-4\times12\times1}}{2\times12}=\frac{24\pm\sqrt{576 - 48}}{24}=\frac{24\pm\sqrt{528}}{24}=\frac{24\pm4\sqrt{33}}{24}=1\pm\frac{\sqrt{33}}{6}\approx1\pm0.957$ The critical points are $x_1=1 - 0.957=0.043$ and $x_2=1 + 0.957 = 1.957$ ## Step2: Test intervals - For $x\in(-\infty,0.043)$, let $x = 0$, then $P^{\prime}(0)=-2<0$, so $P(x)$ is decreasing on $(-\infty,0.043)$ - For $x\in(0.043,1.957)$, let $x = 1$, then $P^{\prime}(1)=-24 + 48-2=22>0$, so $P(x)$ is increasing on $(0.043,1.957)$ - For $x\in(1.957,\infty)$, let $x = 2$, then $P^{\prime}(2)=-24\times4+48\times2-2=-96 + 96-2=-2<0$, so $P(x)$ is decreasing on $(1.957,\infty)$ # Answer: $P(x)$ is increasing on $(0.043,1.957)$ and decreasing on $(-\infty,0.043)\cup(1.957,\infty)$ ### f) # Explanation: ## Step1: Find the optimal production level Set $P^{\prime}(x)=0$, from part (e) $x = 1\pm\frac{\sqrt{33}}{6}$. We take $x = 1+\frac{\sqrt{33}}{6}\approx1.96$ (since $x>0$ for production). ## Step2: Calculate the profit Substitute $x = 1+\frac{\sqrt{33}}{6}$ into $P(x)=-8x^{3}+24x^{2}-2x$ $P(1.96)=-8\times(1.96)^{3}+24\times(1.96)^{2}-2\times(1.96)$ $P(1.96)=-8\times7.535+24\times3.842-3.92$ $P(1.96)=-60.28+92.21 - 3.92=28.01$ (in hundreds of JPY) # Answer: The optimal production level is approximately $196$ rackets ($x\approx1.96$ in hundreds), and the expected profit is approximately $2801$ JPY ($P(x)\approx28.01$ in hundreds of JPY)

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QUESTION IMAGE

Question

Explanation:

a)

Step1: Differentiate term - by - term

Step1: Substitute \(x = 1\) into \(C^{\prime}(x)\)

Step2: Interpret the result

Step1: Use the profit formula \(P(x)=R(x)-C(x)\)

Answer:

b)