. Given the total utility function (T u) = 150x+40x2 = x 3 then Derive marginal utility function and find the value of x at which total utility is maximum
"MU = TU'(x) = 150 + 80x - 3x^2."
Total utility is maximized, if MU = 0, so:
"150 + 80x - 3x^2 = 0,"
D = 8,200,
"x1 = \\frac{-80 + 8,200^{0.5}} {2\u00d73} = 1.76."
x2 < 0, so is not suitable for our case.
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