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?
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."