Alfredo lives on apples and bananas only.His utility function is U(a, b)  mina  b, 2b.He maximizes his utility subject to his budget constraint and consumes the bundle (a, b)  (4, 4) .

A) p_a  p_b.
B) p_a is less than or equal to p_b.
C) p_a  p_b.
D) p_a  2p_b.
E) None of the above.

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Tamara Johnson · Accepted Answer

Final Answer : B, Explanation :  To solve this problem, we need to use the marginal rate of substitution (MRS) and the prices of apples and bananas.The MRS is the amount of bananas that Alfredo is willing to give up for an extra apple, while still maintaining the same level of utility. In this case, the MRS is 2, since Alfredo is willing to give up 2 bananas for an extra apple, while still getting the same level of utility.The prices of apples and bananas are not given, so let's call them pA and pB respectively. Alfredo's budget constraint can be written as pA * a + pB * b = I, where I is his income. Since we don't know the actual prices, let's assume that pA = pB = 1. Then, his budget constraint becomes a + b = I, where I is his income.Alfredo's optimal consumption bundle is the point where the MRS equals the price ratio (pA/pB). In this case, the MRS is 2, so the optimal consumption bundle is the point where a/b = 2/1, or a = 2b.Since Alfredo is currently consuming (4, 4), we know that a/b is currently 1. This means that Alfredo is currently consuming less apples and more bananas than he should be. In order to optimize his consumption, he should decrease his consumption of bananas and increase his consumption of apples.Therefore, the correct answer is B: p a  is less than or equal to p b . This is because the price of apples (pA) should be lower than the price of bananas (pB), since the optimal consumption bundle is at a/b = 2/1.

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Alfredo lives on apples and bananas only.His utility function is U(a, b)  mina  b, 2b.He maximizes his utility subject to his budget constraint and consumes the bundle (a, b)  (4, 4) .A) pa  pb.B) pa is less than or equal to pb.C) pa  pb.D) pa  2pb.E) None of the above.

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