Variation problems involve formulas which show the relationship between two or more variables. A relationship between variables is called joint variation when one variable varies directly with the product of two or more variables. An example of joint variation is the formula for the area of a triangle. Recall that the area of a triangle is ½ the base times the height. (A = (1/2)bh) In general, if y varies jointly with x and z, then y = kxz, where k is the constant of variation and k > 0.

**Example:** Suppose y varies jointly with x and z and y is 12 when x is 3 and z is 2. What is the value of y when x is 8 and z is 5 ?

Start with the equation y = kxz and substitute 12 for y, 3 for x and 2 for z and solve for k.

12 = k(3)(2)

12 = 6k

2 = k

Now substitute 2 for k, 8 for x and 5 for z in the original equation and solve for y.

y = (2)(8)(5)

y = 16(5)

y = 80

**Example:** Suppose the costs incurred by a lumber hauling company varies jointly with the number of trucks used to haul the lumber and the number of hours each truck is used. The cost is $3,400 when 6 trucks are used for 5 hours each. What are the costs when 8 trucks are used for 7 hours each ?

Let c = costs, n = number of trucks and h = hours each truck is used. Therefore the equation used is c = khn. Substitute $3,400 for c, 5 for h and 6 for n. Then solve for k.

3,400 = k(5)(6)

3,400 = 30k

113 1/3 = k

Now substitute 113 1/3 for k, 7 for h and 6 for n in the original equation and solve for c.

c = (113 1/3)(7)(6)

c = (113 1/3)(42)

c = $4, 760 (multiply on a calculator or change 113 1/3 to 340/3 and multiply by 42 to get (340/3)(42) = 340(14) = $4,760)

**Example:** The cost of concrete used to pave a sidewalk varies jointly with the length, width and depth of each rectangular piece. If a piece costs $300 for a piece 5 feet long, 3 feet wide and 6 inches thick, how much would it cost for a piece 12 feet long, 4 feet wide and 6 inches thick ?

Let C = cost, l = length, w = width and d = depth. Use k for the constant of variation.

C = klwd

Substitute the values for C, l, w and d and solve for k.

300 = k(5)(3)(1/2)

300 = (15/2)k

40 = k

Now substitute 40 for k, 12 for l, 4 for w and ½ for d into the original equation and solve for C.

C = 40(12)(4)(1/2)

C = $960

Another method to solve a problem like this is to get the volume of each piece and calculate how many times larger the one piece is. Then multiply that value by $300. The volume of the second piece is (12)(4)(1/2) = 24 feet3 . The volume of the first piece is (5)(3)(1/2) = 7.5 feet3 . The volume of the second one is 24/(7.5) = 3.2 times the volume of the first, so the cost is 3.2 times the cost of the first. Therefore 300(3.2) = $960.

Essentially, the problem was solved by setting up a proportion.

Think of the proportion as

Cost 1/Volume 1 = Cost 2/Volume 2

300/(15/2) = C/24

40 = C/24

C = 960.

This guide should assist anyone having difficulty understanding the concept of joint variation.