Question

When they invest K dollars, the fishery can produce Q(K) pounds of fish per month, where Q(K) = 106K⁽¹/³⁾ months from now. Their total investment will be K(t) = -0.5t² + 200504 dollars.

At what rate is the production of fish changing with respect to time, months from now?

Answer 1

To find the rate of change of the fish production with respect to time, we need to find the derivatives of the fish production function Q(K) and the investment function K(t). Then, we can multiply these derivatives together to calculate the rate of change.

Step-by-step explanation:

The production of fish is changing with respect to time, given by the derivative of the fish production function Q(K). To find this rate of change, we need to take the derivative of Q(K) with respect to K and then multiply it by the derivative of K(t) with respect to t.

First, let's find the derivative of Q(K):

Q'(K) = (1/3) * (106 / K^(2/3))

Now, let's find the derivative of K(t):

K'(t) = -t

Finally, we can find the rate of change of the fish production with respect to time:

dQ/dt = Q'(K) * K'(t)

Substituting in the derivatives we found:

dQ/dt = (1/3) * (106 / K^(2/3)) * (-t)