Question

Maximising revenue subject to minimal profit

A startup maximises revenue q*a from quantity q ≥ 0 and advertising a ≥ 0 subject to a bankruptcy constraint that the profit π = 3*q*a -q^3 -a^3 is greater than m=0.50. Here, ^ denotes power, * multiplication, / division, + addition, - subtraction.

Find the total derivative of the maximised revenue w.r.t. the minimal required profit m at m=0.50. This derivative equals a Lagrange multiplier. Write the answer as a number in decimal notation with at least two digits after the decimal point. No fractions, spaces or other symbols.

Answer #1

Revenue, R = qa

Profit, π = 3qa - q^{3}-a^{3}

π > m = 0.5

So, 3qa - q^{3}-a^{3} > 0.5 ---- (1)

Using Lagrange's multiplier:

L = qa + λ(3qa - q^{3}-a^{3}-0.5)

Differnetiating:

On solving (2) and (3):

q =a

Substituting this in 4 we get:

**q = 1.366, 0.5, -0.366**

**q** cannot be negative so q = 1.366 or 0.5

Substituting q = 1.366 on (1)

we get 0.51 <= 0.5, Hence rejected.

**Substituting q = 0.5 on (1)**

**we get 0.5 <= 0.5, which is admissible**

**Therefore q = 0.5, a =0.5**

Hence, revenue is maximum at q = 0.50 and a= 0.50

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