Mathematics: 0) Assume the total revenue from the sale of x items is given by R(x) = 31 ln (5x + 1), while the total cost to produce x items is C(x) = x 3 . Find the approximate number of items that should be manufactured so that profit, P(x), is maximum

0) Assume the total revenue from the sale of x

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23.01.2022

93 items should be manufacture so that the profit is maximum.

Step-by-step explanation:

Given that,

Total revenue from x items is

R(x)= 31 in (5x+1)

Total cost to produce x items is

$C(x)=\frac x3$

Profit = revenue- cost

P(x)=31 ln (5x+1) - $\frac x3$

Differentiating with respect to x

$P'(x)= 31 \frac{5}{(5x+1)}-\frac13$

Again differentiating with respect to x

$P''(x)= -31 \frac{25}{(5x+1)^2}$

First, we set P'(x) =0

$31 \frac{5}{(5x+1)}-\frac13=0$

$\Rightarrow \frac{155}{(5x+1)}-\frac13=0$

$\Rightarrow \frac{155\times 3-1.(5x+1)}{3(5x+1)}=0$

⇒465 - 5x-1 =0

⇒5x= 464

$\Rightarrow x=\frac{464}{5}$

$\Rightarrow x=92.8$

$\Rightarrow x\approx 93$

Now $P''(x)|_{x=93}= -31 \frac{25}{(5\times 93+1)^2}$

So, at x=93 the profit is maximum.

93 items should be manufacture so that the profit is maximum.

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