# 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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from General Studies Math: Help & Review

Chapter 5 / Lesson 2