Multiple parts Calculus 1 question (easy.. but fully explain in neat handwriting

Question

Multiple parts Calculus 1 question (easy.. but fully explain in neat handwriting.. or typing.. no cursive) Every part as if teaching to someone who doesn't know calculus

Suppose you are attempting to preserve a rare fish population in a remote lake in the wilderness. You have found that by adding fish food to the lake you can increase or decrease the population of rare fish accoring to the equation

where x represents the weight (kilograms) of fish food you add to the lake and f(x) measures the change in the rare fish population after one year [f(x) > 0 means the fish population increeases, f(x) < 0 means the fish population decreases]

a) Would it be a good idea to add as much fish food as possible?

b) Use first and second derivative to find all critical and inflection points of f(x)

to describe the effect of fish food x on a rare fish population.

c) What feeding strategy (what value of x) will increase the rare fish population the most?

HINT: Use calculs and your answers to part (b) to locate and clarify extreme values of f(x)].

Explanation / Answer

SOLUTION

Since all the questions are centered around derivative, we will first find the derivatives and then interpret the results to answer the questions.

Back-up Theory

1. Standard formulae: derivative of xn = nxn – 1 and derivative of ky = ktimes derivative of y where k is a constant

2. To find the critical points including points of inflexion, set the first derivative to zero and solve for x. For each of the values of x obtained as above, find the value of the second derivative. If this value is positive, the corresponding x-value is the minima point, if negative, maxima point and if zero, point of inflexion. Substituting these values of x in the given function, its maximum and minimum values are obtained.

3. Second derivative, denoted by (d2y/dx2) is obtained by differentiating the first derivative

Just for convenience, let us replace f(x) by y. i.e., y = -(x3/3) + 3x2– 5x ……. (1)

Differentiating (1) w.r.t x,

(dy/dx) = - (3x2/3) + 3(2x) – 5   

Simplifying, (dy/dx) = - x2 + 6x – 5 ……… (2)

Setting (dy/dx) to 0, - x2 + 6x – 5 = 0 or x2 - 6x + 5 = 0 or (x - 5)(x - 1) = 0

=> x = 5 or 1…… (3)

Differentiating (2) w.r.t x, (d2y/dx2) = - 2x + 6 …… (4)

Substituting (3) in (4): second derivative is negative at x = 5 and positive at x = 1.

=> y is maximum at x = 5 and minimum at x = 1.

Substituting (3) in (1): ymax = 25/3 and ymin = - 7/3 ……. (5)

Now to interpret and answer the questions,

(5) => maximum fish production possible is 25/3 at feed of 5 and hence it would not serve any purpose adding feed beyond 5 . ANSWERS part (a)

(3) => critical points are x = 5 and x = 1. There are no points of inflexion ANSWERS part (b)

(5) => maximum fish production possible is 25/3 at feed of 5 ANSWERS part (c)

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