In a species of fish, the growth rate function is given by G(x)=0.5x(1-x/K) where K=6000 metric...

Question:

In a species of fish, the growth rate function is given by {eq}G(x)=0.5x(\frac{1-x}{K}) {/eq} where K=6000 metric tons ( the population of fish is measured in metric tons rather than the number of individuals). The price a fisher can get is p=600$ per metric ton. If the amount the fisher can harvest is determined by the function H=hx, where each unit of h costs c=250$, what is the maximum amount of money the fisher can expect to make on a sustainable basis? (Hint: the fisher's sustainable income is given by pH-ch, where H is a sustainable harvesting rate)

Maximum and minimum of given function:


Consider a function {eq}f(x) {/eq} which is continuous and differentiable on the given interval {eq}[a,b]. {/eq} Then the function has its minimum and maximum value at the critical point which we get by differentiating the given function and then equate that derivative with zero.


Answer and Explanation:


The final profit function of the fisherman is given by:

{eq}\hspace{30mm} \displaystyle{ P(x) = pH - ch \\ P(x) = pH - \dfrac{cH}{x} \\ P(x) = p G(x) - \dfrac{c G(x)}{x} \\ P(x) = G(x) \left( p - \dfrac{c}{x} \right) \\ P(x) = \dfrac{1}{2} x \left( 1 - \dfrac{x}{6000} \right) \left( 600 - \dfrac{250}{x} \right) \\ P(x) = \dfrac{\left(-x+6000\right)\left(12x-5\right)}{240} \\ P(x) = \dfrac{ -12x^2 + 72005x - 30000 }{240} \\ } {/eq}

Now we have to find the value of {eq}x {/eq} for which the function {eq}P(x) {/eq} has the maximum value.

{eq}\hspace{30mm} \displaystyle{ P'(x) = \dfrac{-24x + 72005}{240} \\ P'(x) = 0 \implies x = \dfrac{72005}{24} } {/eq}

And {eq}P''(x) = - \dfrac{1}{10} < 0. {/eq} Therefore the function {eq}P(x) {/eq} has the maximum value at {eq}x = \dfrac{72005}{24}. {/eq}

Therefore the maximum amount of money the fisher can make is :

{eq}\hspace{30mm} \displaystyle{ P \left( \dfrac{72005}{24} \right) \approx $ \ 449937.50 } {/eq}



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