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The production function is $f(L,M)=4L^{0.5}M^{0.5}$ , where L is the number of units of labor and M is the number of machines used . If the cost of labor is \$100 per unit and the cost of machines is \$16 per unit, then the total cost of producing 7 units of output will be:

a. \$140. b. \$406. c. \$112. d. \$280. e. \$ None of the above.

The answer is a). Why is that? I am not able to come up with any explanation it is not simply isoquant slope equal isocost slope, thus how to solve this question?

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You need to produce seven units of output, therefore: $7 = 4L^{0.5}M^{0.5}$. Squaring, you find that are restricting yourself to the curve $LM = 49/16$ in the first quadrant as $L,M>0$.

The cost function is $C(L,M) = 100L + 16M$. You want to minimize this subject to the constraint $LM = 49/16$. We can do this more generally, but in this case, it is simple enough to solve $L = \frac{49}{16M}$ and so $$ C = 100\frac{49}{16M} + 16M. $$

Now we have a function defined on $(0,\infty)$. Notice that $\lim_{M\to 0}C = \lim_{M\to\infty}C = \infty$, so the function has a global minimum. As it is differentiable, that global minimum will be at a critical point. So find the critical points of $C$ (i.e., solve $C' = 0$) and figure out which has the lowest value of $C$.


If you take a step back, notice that by the multivariable chain rule, denoting a derivative by subscripting and using the implicit function theorem, finding critical points involves solving the equation $$ C'(L(M),M) = C_L(L,M)L_M + C_M(L,M) = 0 $$ Note that $-C_M/C_L$ is the slope of the isocost curve.

What is $L_M$? This is defined implicitly: $\frac{\partial}{\partial M}f(L(M),M) = f_L(L(M),M)L_M + f_M(L(M),M)$ so $L_M = -f_M/f_L$. This is the slope of the isoquant curve.

Solving, the, we have $L_M = -C_M/C_L$. In other words, the condition that $C' = 0$ boils down to "the slope of the isocost curve equals the slope of the isoquant curve."

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