Question #142339

38. The McNight Company is a major producer of steel. Management estimates that the demand for the company’s steel is given by the equation:

Qs = 5,000 – 1,000Ps + 0.1I + 100Pa

Where Qs is steel demand in thousands of tons per year, Ps is the price of steel in dollars per pound, I is income per capita, and Pa is the price of aluminum in dollars per pound. Initially, the price of steel is $1 per pound, income per capita is $20,000, and the price of aluminum is $0.80 per pound.

(a) How much steel will be demanded at the initial prices and income?

(b) What is the point income elasticity at the initial values?

(c) What is the point cross elasticity between steel and aluminum? Are steel and aluminum substitute or complements?

(d) If the objective is to maintain the quantity of steel demanded as computed in part (a), what reduction in steel prices will be necessary to compensate for a $0.20 reduction in the price of aluminum?

Qs = 5,000 – 1,000Ps + 0.1I + 100Pa

Where Qs is steel demand in thousands of tons per year, Ps is the price of steel in dollars per pound, I is income per capita, and Pa is the price of aluminum in dollars per pound. Initially, the price of steel is $1 per pound, income per capita is $20,000, and the price of aluminum is $0.80 per pound.

(a) How much steel will be demanded at the initial prices and income?

(b) What is the point income elasticity at the initial values?

(c) What is the point cross elasticity between steel and aluminum? Are steel and aluminum substitute or complements?

(d) If the objective is to maintain the quantity of steel demanded as computed in part (a), what reduction in steel prices will be necessary to compensate for a $0.20 reduction in the price of aluminum?

Expert's answer

a) "Q_{s}=5000-1000\\times1+0.1\\times20000+1000\\times0.80=6080"

(b) Point income elasticity ="\\frac{dQ_{d}}{dI}\\times \\frac{I}{Q_{d}}"

="0.1\\times20000\/6080=0.33"

(c) point cross elasticity = "\\frac{dQ_{s}}{dP_{a}}\\times \\frac{P_{a}}{Q_{s}}"

="100\\times 0.80\/6080 = 0.01"

(d) "6080 = 5000-1000\\times P_{s} + 0.1\\times20000+100\\times 0.6"

"1000P_{s}=980"

"P_{s} = 0.98"

required reduction = "\\frac{(1-0.98)}{1}\\times 100 = 2\\%"

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