Question

Suppose the manufacturer can sell X items per week at a price , p=(20-0.001X) rupees each when it costes, Y=5X+2000 rupees to produce X items. Determine the number of items to be produced & week for maximum profit?​

Answer 1

Answer:

SSam earned 9870 rupee the last month. He spent 3775 rupees on food 1050 rupees more on other items them on food and save the rest. How much did he saved.​tep-by-step explanation:

Answer 2

Answer:

Given:

The manufacturer sells X items per week at a price of (20-0.001X) Rupees.

Therefore the total revenue obtained would be,

                      $= x(20-0.001x) = 20x-0.001x^2$

The cost price of x items is also given. The cost price of x items = 5x+2000

Let P(x) be the function for profit.

Using the formula, Profit = Revenue - cost

                     $\therefore \quad P=\left(x(20-0.001x\right))-(5x+2000)$

                                $=20x-0.01x^2-5x-2000$

                               $=-0.01x^2+15x-2000$

Now differentiate P w.r.t x ,

                                  $\frac{d P}{d x}=-0.02x+15$

To find maxima and minima, put  $\frac{d P}{d x}=0$ or $-0.02x+15=0$  or x = 750

Also,

$\frac{d^{2} P}{d x^{2}}=\frac{d}{d x}\left(\frac{d P}{d x}\right)=\frac{d}{d x}(-0.02x+15)$

                        = $-0.02 < 0$

Thus, at $x=750, \frac{d^{2} P}{d x^{2}} < 0 or P is maximum.$

$=\left[x\left(5-\frac{x}{100}\right)\right]=\left(5 x-\frac{x^{2}}{100}\right)$

Hence, the number of items sold to have maximum profit is 750.

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