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 - q³-a³

π > m = 0.5

So, 3qa - q³-a³ > 0.5 ---- (1)

Using Lagrange's multiplier:

L = qa + λ(3qa - q³-a³-0.5)

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

**Hope this helped. Please Upvote**