In a company which services air-conditioner units, the revenue,
$r, is given by the equation r = 20x, where x is the number of
units serviced. The operating cost, $c, is given by the equation
c= 3600 + 4x.
The profit, $P, is given by the expression, r — c.
(a) Find a formula for P in terms of x only.
(b) For each of the following cases, find the value of x if
(i) the revenue earned is twice that of the cost,
(ii)the profit earned is $292 more than one third of the revenue
earned.

The problem statement gives expressions and relations for revenue, cost, and profit. Substitute and simplify or solve, as appropriate.

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(a)

Given: r(x) = 20x; c(x) = 3600+4x; p(x) = r(x) -c(x)

The formula for p(x) is ...

p(x) = (20x) -(3600 +4x) . . . . substitute for r(x) and c(x)

p(x) = 16x -3600 . . . . . . . . simplify

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(b-i)

r(x) = 2·c(x) . . . . given relation for revenue and cost

20x = 2(3600 +4x) . . . . substitute for each

12x = 7200 . . . . . . . simplify and subtract 8x

x = 600 . . . . . . . divide by 12

Revenue is twice the cost when x = 600 units.

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(b-ii)

p(x) = 292 +1/3r(x) . . . . given relation for profit and revenue

16x -3600 = 292 +1/3(20x) . . . . . substitute for each

28/3x = 3892 . . . . . . . . add 3600 -20/3x

x = 417 . . . . . . . . multiply by 3/28

Profit is $292 more than 1/3 the revenue earned for 417 units.