(1 point) A bird in its natural habitat feeds on two kinds of seeds, whose nutri

Question

(1 point) A bird in its natural habitat feeds on two kinds of seeds, whose nutritional values are 5 calories per seed of type 1 and 3 calories per seed of type 2. Both kinds of seeds are hidden among litter on the forest floor and have to be found. If the bird splits attention into a fraction x for food type 1 and x2 for food type 2, then its probability of finding a seed of the given type is P,(xl ) = (xl )3, P2(x2 ) = (x2 )5 Assume that the bird pays full attention to searching for seeds so that x + x2-1 where 0 x1S 1 and 0 x2 1 . The expected nutritional value of any seed found on the forest floor can be calculated by multiplying the nutritional value of a seed type by the probability of finding that seed type and adding these products up over all seed types. (a) Write down an expression for the expected nutritional value V(x) gained by the bird when it splits its attention. Use the constraint on x to eliminate x2 and rename the variable xi as x. V(x) = (b) Find the critical point of Vx) Is this critical point a local minimum or maximum? A. Local maximum. B. Local minimum. C. Neither. (c) Find absolute minima and maxima of V(x) Absolute minima-F Absolute maxima Use your results to determine the optimal strategy of the bird to maximize the nutritional value of the seeds it can find. A. The bird should spend all its attention on food of type 1 and none on food of type 2 to get the most nutritional value. B. The bird should spend its attention on a combination of both types of food to get the most nutritional value C. The bird should spend all its attention on food of type 2 and none on food of type 1 to get the most nutritional value.

Explanation / Answer

Here x1 + x2 =1 where 0 < x1,x2 < 1

Expected Nutrition value E(NV) P = 5 P1(x1) + 3 P2(x2)

5 * x13 + 3 * x25 = 5x13 + 3x25

(b) V(x) = 5x13 + 3x25

V(x) =5x13 + 3 (1-x1)5 = 5x3 + 3(1-x)5

dV(x)/ dx = 15 x2  + 15 (1-x)4

x2= (1 - x)4

x= +- (1 - x)2

+-x = (1-x)2

+-x = 1 + x2 -2x

x2-3x +1 = 0; x2 -x +1 =0

x = 1 - 0.618 = 0.382

V(x)critical = 5 * (0.382)3 + 3 * (0.612)5= 0.5363

This point is critical minima. as d(V)/dx is positve

(c) Absolute minima = 0.5363

Absolute minima would be for x1 =1 and x2 = 0

Absolute Maxima = 5

Here the bird shoudl concentrate all its attention on food of type 1 and none on food of type 2 to get the most nutrititonal value. Option A is correct.

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