uations 2xz + xy + z – 2z = 11 xyz = 6 satisfied. One solution of this set of equations is x = rthern Saskatchewan is in equilibrium at this point. Suppose that the prim nister discovers that the variable z (output of beaver pelts) can be conrolled bị 3, у %3D 2,z 1, an %3D

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Could you please explain the following question, the textbook gave the answer below, but I am not sure how they get there. Q.a?

15.21 The economy of Northern Saskatchewan is in equilibrium when the system of
equations
2xz + xy + z – 2/z = 11
xyz = 6
is satisfied. One solution of this set of equations is x =
Northern Saskatchewan is in equilibrium at this point. Suppose that the prime
minister discovers that the variable z (output of beaver pelts) can be conrolled by
simple decree.
a) If the prime minister raises z to 1.1, use calculus to estimate the change in x and
3, у
2, z
1, and
у.
b) If x were in the control of the prime minister and not y or z, explain why you
cannot use this method to estimate the effect of reducing x from 3 to 2.95.
Transcribed Image Text:15.21 The economy of Northern Saskatchewan is in equilibrium when the system of equations 2xz + xy + z – 2/z = 11 xyz = 6 is satisfied. One solution of this set of equations is x = Northern Saskatchewan is in equilibrium at this point. Suppose that the prime minister discovers that the variable z (output of beaver pelts) can be conrolled by simple decree. a) If the prime minister raises z to 1.1, use calculus to estimate the change in x and 3, у 2, z 1, and у. b) If x were in the control of the prime minister and not y or z, explain why you cannot use this method to estimate the effect of reducing x from 3 to 2.95.
15.21 a)
(2z + y) dx + x dy + (2x + 1 –z-1/2) dz = 0
(yz) dx + (xz) dy + (xy)
dz = 0,
74
MATHEMATICS FOR ECONOMISTS
or
4 dx + 3 dy + 6(0.1) = 0
2 dx + 3 dy + 6(0.1) = 0.
O and dy
Subtracting yields dx
y = 1.8.
-0.6/3
— —0.2. So, х ~ 3,
Transcribed Image Text:15.21 a) (2z + y) dx + x dy + (2x + 1 –z-1/2) dz = 0 (yz) dx + (xz) dy + (xy) dz = 0, 74 MATHEMATICS FOR ECONOMISTS or 4 dx + 3 dy + 6(0.1) = 0 2 dx + 3 dy + 6(0.1) = 0. O and dy Subtracting yields dx y = 1.8. -0.6/3 — —0.2. So, х ~ 3,
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