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# Maximize each of the following utility functions, with the cost of each commodity and total amount available to spend given. $f(x, y)=x^{3} y^{4}$, cost of a unit of is $3, cost of a unit of is$3, and \$42 is available.

Solution

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Once again we use the fact that neither $x$ or $y$ can be zero if we are trying to obtain the maximum. The constraint function is $g(x,y)=3x+3y=42$. Lagrange multiplier equations are:

\begin{align*} &3x^2y^4=3\lambda\\ &4x^3y^3=3\lambda\\ &x+y=16\\ \end{align*}

Combining the first and the second equations give us $x^2y^3(4x-3y)=0$. Use the third equation to get $x=6$. From there $y=10$. Hence the maximum is at $(6,10)$.

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