Ambrose's utility function is  4x^1/2₁  x₂.If the price of nuts (good 1) is $1, the price of berries (good 2) is $9, and his income is $369, how many units of berries will Ambrose choose?

A) 10
B) 4
C) 325
D) 5
E) 8

We need to maximize Ambrose's utility given his budget constraint, which is:
$1F_1 + $9F_2 ≤ $369
We can rearrange this to:
F_2 ≤ ($369 - $1F_1)/$9
Now let's look at the utility function. We can simplify it to:
F_1^(1/2)S_1^(1/2) + 4F_1^(1/2)U_1P_1^(1/2) + S_1U_1B_1^(1/2) + F_2^(1/2)S_1^(1/2) + F_2^(1/2)U_1B_1^(1/2)
The first term represents the utility from consuming nuts, while the second term represents the disutility from their high price (U_1 is a disutility function). The third term represents the utility from consuming berries, while the fourth and fifth terms represent the disutilities from their high price and from the fact that they require effort to collect (B_1 is an effort function).
To maximize utility, we need to find the optimal combination of F_1 and F_2 that satisfies the budget constraint. We can do this using Lagrange multipliers. The Lagrangian is:
L = F_1^(1/2)S_1^(1/2) + 4F_1^(1/2)U_1P_1^(1/2) + S_1U_1B_1^(1/2) + F_2^(1/2)S_1^(1/2) + F_2^(1/2)U_1B_1^(1/2) - λ($1F_1 + $9F_2 - $369)
Taking partial derivatives and setting them equal to zero, we get:
(1/2)F_1^(-1/2)S_1^(1/2) + 2(1/2)F_1^(-1/2)U_1P_1^(1/2) - λ = 0
(1/2)F_2^(-1/2)S_1^(1/2) + (1/2)F_2^(-1/2)U_1B_1^(1/2) - 9λ = 0
$1F_1 + $9F_2 = $369
Solving for λ in the second equation and substituting into the first equation, we get:
(1/2)F_1^(-1/2)S_1^(1/2) + 2(1/2)F_1^(-1/2)U_1P_1^(1/2) = (1/18)(1/2)F_2^(-1/2)S_1^(1/2) + (1/18)(1/2)F_2^(-1/2)U_1B_1^(1/2)
Simplifying, we get:
F_1^(1/2) = (1/9)F_2^(1/2)
Substituting into the budget constraint, we get:
$1(1/9^(1/2))F_2^(1/2) + $9F_2 = $369
Solving for F_2, we get F_2 = 20.25
Therefore, Ambrose will choose $9F_2 = $182.25 worth of berries, or about 20 berries (since each berry costs $9). The answer is D.