Average utility;
Lets assume number of units consumed is "x"
Average utility=total utility"\\div" no of units consumed
="\\frac{150x+40x^2-x^3}{x}"
Marginal utility;
Marginal utility, MU="\\frac{\\Delta u}{\\Delta x}=150+80x-3x^2"
Total utility;
Total utility is maximized when MU=0
"150+80x-3x^2=0"
D = 8,200
"x_{1} = \\frac{-80 + 8,200^{0.5}} {2\u00d73} = 1.76."
"x_{2}" < 0, so is not suitable for our case
Value of x at which average utility is maximum;
="\\frac{150x+40x^2-x^3}{x}"
"150+40x-x^2=0"
"x=\\frac{\u221240\u00b1\u221a2200}{\u22122}"
"x_{1}=20\u22125\u221a22 =-3.45"
"x_{2}=20+5\u221a22=43.45"
Our value will be 43.45
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