Dimishing returns weight loss word problem help

Question

Mike was successful in losing weight. He had target weight in mind. He went on diet for 3 months and each month, he would lose 1/3 of difference between his current weight and target weight plus additional 3 pounds. At the end of 3 months, he was 3 pounds over his target weight. How many pounds did he lose in 3 months?How do you solve this algebraically using equations? Please help me.

Answer 1

Kaito,

Given the way that Mike's monthly weight loss is defined ("difference between his current weight and target weight"), it helps to focus on his weight w at the start of each month--his "current weight".

In (at the end of) the first month, Mike will have lost w(1) - w(2) = [w(1) - target]/3 + 3 > 0 pounds during that month where:

w(1) = weight at start of month 1

w(2) = weight at start of month 2 = w(1) + [w(1) - target]/3 + 3 = 2w(1)/3 + target/3 - 3

w(3) = weight at start of month 3 = w(2) + [w(2) - target]/3 + 3 = 2w(2)/3 + target/3 - 3

w(4) = weight at start of month 4 = w(3) + [w(3) - target]/3 + 3 = 2w(3)/3 + target/3 - 3

target = w(4) - 3

Note that w(1) - w(2) = [w(1) - target]/3 + 3 can be rearranged to be w(2) = w(1) - [w(1) - target]/3 - 3

or w(2) = (2/3)w(1) + (1/3)(target - 9), which makes perfect sense when you think about what Mike's weight is at the start of the second month: He has lost a third of his original weight relative to a third of his target weight plus three pounds.

We want an expression for the weight Mike lost in three months w(4) - w(1), so we start with the expression for w(4) above and substitute successively for w(3) and for w(2):

w(4) = (2/3)w(3) + (1/3)(target - 9)

w(4) = (2/3) [(2/3)w(2) + (1/3)(target - 9)] + (1/3)(target - 9)

w(4) = (2/3)^2 w(2) + (2/3)(1/3)(target - 9) + (3/3)(1/3)(target - 9)

w(4) = (2/3)^2 w(2) + 5(1/3)^2 (target - 9)

w(4) = (2/3)^2 [(2/3)w(1) + (1/3)(target - 9)] + 5(1/3)^2 (target - 9)

w(4) = (2/3)^3 w(1) + (2/3)^2 (1/3)(target - 9) + 5(3/3)(1/3)^2 (target - 9)

w(4) = (2/3)^3 w(1) + 19(1/3)^3 (target - 9)

Subtract w(1) from this expression for w(4) to get Mike's total weight loss.

w(i) = (2/3)w(i - 1) + (1/3)(target - 9) is an example of a "difference equation".

It can be solved by successive substitution as above,

but there are also formal methods that will produce a simplified expression for any value of w(i).

Note that the solution involves powers of the various coefficients, such as (2/3)^3.

This is characteristic of the solutions of difference equations.

Note also that I kept (target - 9) together with a coefficient because both are simply constants.

Please let me know should you have any questions. Thanks.

Michael (a.k.a. Mike)