Algebra Demystified 2nd Ed

234 algebra De mystif ieD

EXAMPLE

The cost formula for a manufacturer’s product is C = 5000 + 2x, where C

is the cost (in dollars) and x is the number of units manufactured.

(a) If no units are produced, what is the cost?

(b) If the manufacturer produces 3000 units, what is the cost?

(c) If the manufacturer has spent $16,000 on production, how many units

were produced?

We answer these questions by substituting the numbers into the formula.

(a) If no units are produced, then x = 0, and C = 5000 + 2 x becomes

C = 5000 + 2(0) = 5000. The cost is $5,000.

(b) If the manufacturer produces 3000 units, then x = 3000, and C = 5000 + 2x

becomes C = 5000 + 2(3000) = 5000 + 6000 = 11,000. The manufac-

turer’s cost is $11,000.

(c) The manufacturer’s cost is $16,000, so C = 16,000. Substitute 16,000

for C in the equation C = 5000 + 2 x, giving us 16,000 = 5000 + 2 x.

16 000 5000 2

5000 5000

11 000 2

11 000

2

5500

,

,

,

=+

−−

=

=

=

x

x

x

xx

There were 5500 units produced.

The profit formula for a manufacturer’s product is P = 2x – 4000 where x is

the number of units sold and P is the profit (in dollars).

(a) What is the profit when 12,000 units were sold?

(b) What is the loss when 1500 units were sold?

(c) How many units must be sold for the manufacturer to have a profit of

$3000?

(d) How many units must be sold for the manufacturer to break even?

(This question could equivalently be phrased, “How many units must be

sold in order for the manufacturer to cover its costs?”)

EXAMPLE

The cost formula for a manufacturer’s product is