Ernie's utility function is U (x,y) = xy. Ernie consumes 4 units of x and 16 units of y.
A: Ernie likes x and y equally well so he is always willing to exchange 1 unit of either good for more than one unit of the other.
B: None of all
C: Ernie would be willing to make small exchanges of x for y in which he gives up 4 units of x for everyunit of y he gets.
D: Ernie would be willing to trade away all of his x for y so long he gets more than 4 units of y for every unit of x he gives up.
A: Ernie likes x and y equally well so he is always willing to exchange 1 unit of either good for more than one unit of the other.
B: None of all
C: Ernie would be willing to make small exchanges of x for y in which he gives up 4 units of x for everyunit of y he gets.
D: Ernie would be willing to trade away all of his x for y so long he gets more than 4 units of y for every unit of x he gives up.
举一反三
- Ike's utility function is U(x;y) = 25xy. He has 12 units of good x and 8 units of y. Ben's utility function for the same two goods is U(x;y)=4x+4y. Ben has 9 units of x and 13 units of y.
- Supposeaconsumerhaspreferencesovertwogoods,XandY,whichareperfectsubstitutes.Inparticular,twounitsofXisequivalenttooneunitofY.IfthepriceofXis$1,thepriceofYis$3,andtheconsumerhas$30ofincometoallocatetothesetwogoods,howmuchofeachgoodshouldtheconsumerpurchasetomaximizesatisfaction?? 0;units;of;X;and;10;units;of;Y|15;units;of;X;and;5;units;of;Y|15;units;of;X;and;0;units;of;Y|30;units;of;X;and;0;units;of;Y
- Matt's utility function is min{x，4y + 5z}. The price of x is 1, the price of y is 4, and the price of z is 7. Matt's income is 8. How many units of x does Matt demand? A: 4 B: 7 C: 3.33 D: 5
- 已知int x=3,y=4;，写出下列表达式的值 (1) (x,y) (2) x>y?x:y (3) x?y:x (4) (x>y)?(y>=2)?1:2:(y>x)?x:y
- 如下C程序的输出是什么？#include [stdio.h]void Func1 (int x, int y);void Func2 (int *x, int *y); int main() { int x = 3; int y = 4;Func1 (x, y); printf ("x = %d, y = %d\n", x, y);Func2(&x, &y); printf ("x = %d, y = %d\n", x, y);} void Func1 (int x, int y) { x = x + y; y = x - y; x = x - y; printf ("x = %d, y = %d\n", x, y);} void Func2 (int *x, int *y) { *x = *x + *y; *y = *x - *y; *x = *x - *y;;} A: x = 3, y = 4x = 3, y = 4x = 3, y = 4 B: x = 4, y = 3x = 4, y = 3x = 4, y = 3 C: x = 3, y = 4x = 3, y = 4x = 4, y = 3 D: x = 4, y = 3x = 3, y = 4x = 4, y = 3