if given the cost function of producing $x$ products as $C(x)=x^3-3x^2-80x+500 $ and each product is sold at $2800$ dollars,what weekly production rate will maximize the profit and what is the largest possible profit per week? here is what i did $R(x)= 2880x$ and $P(x)= R(x)-C(x)$ so $P(x)= -x^3+3x^2+2880x-500$ $P'(x)=-3x^2+6x+2880$ the factors are $32$ and $-30$ so $x$ has zero at $32$ and $-30$ from here,how do I get the maximum point?
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2What did you try? – Martund Dec 29 '19 at 14:44
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i looked for the profit function by taking 2800x-x^3-3x^2-80x+500 and calculated P'(x) to get the max and min points which were-30 and 32.From this I don't know what to do next – Beatrice chakali Dec 29 '19 at 14:47
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2You should include your attempts into the body of your question. – Martund Dec 29 '19 at 14:48
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note: $\pi=TR-TC \Rightarrow \pi'=MR-MC=0$ and $TR=PQ=2800x$. – farruhota Dec 29 '19 at 15:05
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Double differentiate the profit function and check the sign at candidate point to find maxima. We have $$P''(x)=-6x+6$$ Observe that $P''(32)<0$. Hence, $32$ is the required point of maxima. In other words, the optimum number of products to maximise profit is $32$. Hence the required production rate is $32$ products per week.
Martund
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@Beatricechakali, upvote and accept this answer if it solves your problem. – Martund Dec 29 '19 at 15:21
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Substitute the values you got in $P(x)$.
$P'(x)=0$ at a maximum or minimum of point, so substitute both $32$ and $-30$ and the higher is the maximum.
Also note that there can't be $-30$ products!
Another method is to check the double derivative, the value that $P''(x)<0$ on, is the maximum (because the concavity is downward)
Fareed Abi Farraj
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