# x=units sold, Find maximum profit and number of units that must be sold, in order to yield the maximum profit for each R(x)= 20x-0.1x^2, C(x)=4x+2

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The profit is the excess of the sale revenue over the cost.

Therefore,

Profit = (20x-0.1x^2)-(4x+2) = -0.x^2+(20-4)x -2

=-0.1x^2+(20-4)x-2

= -0.1{x^2-16*10x+20)

= -0.1(x^2-160x+20)

=-0.1{(x^2-160x+80^2) - 80^2 +20}. Here 80^2 is added and subtracted to make x^2-160x into (x-80)^2 a complete square.

=-0.1{(x-80)^2 - 6380}. This is negative and so maximum when the quantity inside the bracket is least. That is when x-80 = 0. Or x = 80. And th1 maximum profit is = -(0.1){(x-80)^2-6380} when when x= 0. So max profit = -(0.1){(80-80^2-6380)} = 638.

So we must sell 80 units to get the maximum profit of 638.

Revenue is given by the function:

R(x) = 20x - 0.1x^2

And cost of sales is given by the function:

C(x) = 4x + 2

Then profit will be given by the function:

P(x) = R(x) - C(x) = (20x - 0.1x^2) - (4x + 2)

P(x) = -0.1 x^2 + 16x - 2

To find the value of x where profit is maximum we differentiate P(x) and equate it to 0. Thus

- 0.2x + 16 = 0

0.2x = 16

x = 16/0.2 = 80

Profit when 80 unite are sold:

P(80) = -0.1*80^2 + 16*80 - 2 = - 640 + 1280 - 2 = 638

Answer:

Maximum profit is 638, which will be earned when 80 units are sold.