Maximising revenue subject to minimal profit A startup maximises revenue q*a from quantity 9 2 0 and advertising a 20 subject to a bankruptcy constraint that the profit T1 = 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: Maximising revenue subject to minimal profit A startup maximises revenue q*a from quantity 9 2 0 and advertising a 20 subject to a bankruptcy constraint that the profit T1 = 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