7. A population y changes at a rate modeled by the differential equation
dy /dt= 0.2y (1000 - y), where t is
measured in years. What are all values of y for which the population is increasing at a decreasing rate?
(A) 500 only
(B) 0 < y < 500 only
(C) 500 < y < 1000 only
(D) 0 < y <1000
(E) y > 1000
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We have a population growth problem with the differential equation dy/dt equals 0.2y times 1000 minus y. We need to find when the population is increasing at a decreasing rate. This means the first derivative must be positive and the second derivative must be negative.
First, we need to find when the population is increasing. This happens when dy/dt is greater than zero. So we need 0.2y times 1000 minus y to be positive. Since 0.2 is positive, we need y greater than zero and 1000 minus y greater than zero. This gives us zero less than y less than 1000.
Now we need to find the second derivative. First, we rewrite dy/dt as 200y minus 0.2y squared. Using the chain rule, the second derivative equals the derivative of this expression with respect to y, times dy/dt. The derivative gives us 200 minus 0.4y, so the second derivative is 200 minus 0.4y times 0.2y times 1000 minus y.
For the second derivative to be negative, we need the product of 200 minus 0.4y and 0.2y times 1000 minus y to be negative. Since we already know that 0.2y times 1000 minus y is positive when y is between 0 and 1000, we need 200 minus 0.4y to be negative. This gives us 200 less than 0.4y, which means y greater than 500.
Now we combine both conditions. For the population to be increasing, we need y between 0 and 1000. For the rate to be decreasing, we need y greater than 500. The intersection of these conditions gives us 500 less than y less than 1000. Therefore, the answer is C: 500 less than y less than 1000 only.