A manufacturer offers price per unit based on, order quantity “x” as per following function: `p = 20 – 0.001x` Rs . The cost of manufacturing for X units is given by the function: `C = 5x + 2000` Rs. Determine the number units he should make and sell per week to maximize the profit
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First, determine the revenue (R). To do so, apply the formula:
R=(Price per unit)* (Number of units sold)
Since the price per unit is represented by the function p and x is the number of units, then:
`R=px = (20-0.001x)(x)`
Now that the revenue is known, apply the formula of profit which is:
Profit = Revenue - Cost
Since cost is represented by the function C, then profit (P) is:
`P= R - C`
`P= 20x- 0.001x^2 - (5x+ 2000)`
`P =20x -0.001x^2-5x-2000`
Now that the profit function is known, take the derivative of P to determine the value of x that would maximize the profit.
Then, set P' equal to zero.
And, isolate x.
Hence, to maximize the profit, the manufacturer should produce and sell 7500 units.
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