Formulas in Word Problems Help

Introduction to Formulas in Word Problems - Cost and Profit

For some word problems, nothing more will be required of you than to substitute a given value into a formula, which is either given to you or is readily available. The most difficult part of these problems will be to decide which variable the given quantity will be.

For example, the formula might look like R = 8 q and the value given to you is 440. Is R = 440 or is q = 440? The answer lies in the way the variables are described. In R = 8 q, it might be that R represents revenue (in dollars) and q represents quantity (in units) sold of some item. “If 440 units were sold, what is the revenue?” Here 440 is q. You would then solve R = 8(440). “If the revenue is $440, how many units were sold?” Here 440 is R , and you would solve 440 = 8 q .

Cost Formula Word Problem

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

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 + 2 x becomes C = 5000 + 2(3000) = 5000 + 6000 = 11,000. The manufacturer’s cost would be $11,000.

(c) The manufacturer’s cost is $16,000, so C = 16,000. Substitute C = 16,000 into C = 5000 + 2 x to get 16,000 = 5000 + 2 x .

There were 5500 units produced.

Profit Formula Word Problem

The profit formula for a manufacturer’s product is P = 2 x – 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 have been phrased, “How many units must be sold in order for the manufacturer to cover its costs?”)

(a) If 12,000 units are sold, then x = 12,000. The profit equation then becomes P = 2(12,000) – 4000 = 24,000 – 4000 = 20,000. The profit is $20,000.

(b) Think of a loss as a negative profit. When 1500 units are sold, P = 2 x – 4000 becomes P = 2(1500) – 4000 = 3000 – 4000 = – 1000. The manufacturer loses $1000 when 1500 units are sold.

(c) If the profit is $3000, then P = 3000; P = 2 x – 4000 becomes 3000 = 2 x – 4000.

A total of 3500 units were sold.

(d) The break-even point occurs when the profit is zero, that is when P = 0. Then P = 2 x – 4000 becomes 0 = 2 x – 4000.

The manufacturer must sell 2000 units in order to break even.

Volume Formula Word Problem

A box has a square bottom. The height has not yet been determined, but the bottom is 10 inches by 10 inches. The volume formula is V = lwh , because each of the length and width is 10, lw becomes 10.10 = 100. The formula for the box’s volume is V = 100 h .

(a) If the height of the box is to be 6 inches, what is its volume?

(b) If the volume is to be 450 cubic inches, what should its height be?

(c) If the volume is to be 825 cubic inches, what should its height be?

(a) The height is 6 inches, so h = 6. Then V = 100 h becomes V = 100(6) = 600.

    The box’s volume is 600 cubic inches.

(b) The volume is 450 cubic inches, so V = 450, and V = 100 h becomes 450 = 100 h .

   The box’s height would need to be 4.5 inches.

(c) The volume is 825, so V = 100 h becomes 825 = 100 h .

   The height should be 8.25 inches.

   Suppose a square has a perimeter of 18 cm. What is the length of each of its sides? (Recall the formula for the perimeter of a square: P = 4 l where l is the length of each of its sides.)

   P = 18, so P = 4 l becomes 18 = 4 l .

   The length of each of its sides is 4.5 cm.