Question

Simon spends his income, $100, on food (f) and video games (v) with prices p_f= 7 and p_v = 1. Assume that f ≥ 0, v ≥ 0. His preferences are given by u(f, v) = f^0.7 *v^0.3 . Paul’s Marshallian demand is . (a) (11, 23) (b) (5, 65) (c) (10, 30) (d) (10, 25)

Answer 1

If, we have to maximize the utility function, U(x,y) = x^a y^b; a, b > 0

Subject to : p_x x+ p_y y = m (Budget Line)

For solving this above problem, we have to set Lagrange, L = x^a y^b + ƛ (m - p_x x- p_y y) [ ƛ > 0]

So, differentiate L w.r.t. x we get, dL/dx = a x^{( a – 1 )} y^b – ƛ p_x = 0

Or, ƛ p_x = a x^{( a – 1 )} y^b ................................................(1)

And, differentiate L w.r.t. y we get, dL/dy = b x^a y^{( b – 1 )} – ƛ p_y = 0

Or, ƛ p_y = b x^a y^{( b – 1 )} ................................................(2)

(1)/(2), we get,

p_x/ p_y = ay/bx ................................................(3)

or, x = (ay/b) (p_y/ p_x)

or, y= (bx/ a) ( p_x/ p_y)

Putting equation (3) into the Budget Line, we get,

p_x (ay/b) (p_y/ p_x) + p_y y = m

or, (ay/b) + y = m/ p_y

or, (a + b) y = b m/ p_y

or, y^* = [b/(a + b)] (m/ p_y) ................................................(4)

Similarly putting equation (3) into the Budget Line, we get,

x^* = [a/(a + b)] (m/ p_x) ................................................(5)

Here, m = 100; x = f; p_x = p_f = 7; y = v; p_y = p_v = 1; a = 0.7; b = 0.3

f ≥ 0 and v ≥ 0

So, we have to maximize the utility function, U(f,v) = f^0.7 v^0.3

Subject to : 7 f+ 1 v = 100

So, f^* = [0.7/(0.7 + 0.3)] (100/7) = 10

And, v^* = [0.3/(0.7 + 0.3)] (100/1) = 30

So, Paul’s Marshallian demand is – (c) (10, 30)

So, option (c) is the correct option.

Hope, the explanation is clear to you. Thank you.