Question

Ben swims​ 50,000 yards per week in his practices. Given this amount of​ training, he will swim the​ 100-yard butterfly in 51.5 seconds and place 10th in a big upcoming meet. ​ Ben's coach calculates that if Ben increases his practice to​ 60,000 yards per​ week, his time will decrease to 46.4 seconds and he will place 7th in the meet. If Ben practices​ 70,000 yards per​ week, his time will be 44.6 seconds and he will win the meet. In terms of​ Ben's time in the big​ meet, what is his marginal productivity of the number of yards he​ practices?

Answer 1

Answer:

MArginal productivity: $\frac{dt}{dL}=-0.0002$

We can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

Step-by-step explanation:

The marginal productivy is the instant rate of change in the result for an increase in one unit of a factor.

In this case, the productivity is the time he last in the 100-yard. The factor is the amount of yards he train per week.

The marginal productivity can be expressed as:

$\frac{dt}{dL}$

where dt is the variation in time and dL is the variation in training yards.

We can not derive the function because it is not defined, but we can approximate with the last two points given:

$\frac{dt}{dL}\approx\frac{\Delta t}{\Delta L} =\frac{t_2-t_1}{L_2-L_1}=\frac{44.6-46.4}{70,000-60,000}=\frac{-2.0}{10,000}=-0.0002$

Then we can interpret this as he will reduce his time an additional 0.0002 seconds for every additional yard he trains.

This is an approximation that is valid in the interval of 60,000 to  70,000 yards of training.