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I have no clue how to work with min functions, and am struggling! I have a question that has the constraint of Total Income $=150=I_x+I_y$, thus getting $I_y= 150-I_x$. My function is $W = min (U_x, U_y)$.

My functions are $U_x = 50 \sqrt(I_x)$ and $U_y = 100\sqrt(I_y)$

Subbing in the constraint you get $W = min(50 \sqrt(I_x), 100\sqrt(150-I_x))$.

How do I solve for $I_x$ from here by differentiating and let $W=0$?

In a previous question, once subbing in the constraint into the welfare function, the lecturer differentiated the function and made it $=0$ in order to find the point where welfare is maximised.

Context: This is welfare economics using a Rawlsian social welfare function.

Kyle Britto
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  • May be you can do the following. You can find the point where $U_x=U_y$, thus you find $I_x$. Let we say that the point is $a$. Depending on the interval, i.e. $[0,a]$ and $[a,150]$ (I am assuming that $I_x>=0$ and $I_y>=0$) you can find the minimum of $W$. – tempx Jul 27 '17 at 08:50

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when does $\sqrt{I_x} \leq 2 \sqrt{150-I_x}$?

By squaring both sides, $I_x \leq 4 (150-I_x)$

$$5I_x \leq 600$$

$$I_x \leq 120$$

$$\min\left(50 \sqrt{I_x}, 100 \sqrt{150-I_X} \right)= \begin{cases} 50 \sqrt{I_X} & I_X \leq 120 \\ 100 \sqrt{50-I_X}\ & I_X > 120 \end{cases}$$

We can see that the function increases from $0$ to $120$ and then decreases from $120$ onwards.

Hence the maximum point is at $I_X=120$.

Siong Thye Goh
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    So you essentially are looking for the point in which the two functions intersect? Thank you for that – Kyle Britto Jul 27 '17 at 09:24
  • yup since we understand the function $50\sqrt{x}$ is increasing and $100\sqrt{50-x}$ decreases and they share the same value when they meet. – Siong Thye Goh Jul 27 '17 at 15:15
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The problem is that $W=\min{(U_{x},U_{y})}$ is what is known as a piecewise function, and these are not guaranteed to be differentiable. In this case, you won't be able to differentiate it at the point where $U_{x}=U_{y}$. We'll call that point $a$. To see why, evaluate the derivatives of $U_{x}$ and $U_{y}$ w.r.t. $I_{x}$ at $I_{x}=a$, and you'll find they are unequal. The derivative of $W$ at that point is thus undefined.

There is, in fact, no point at which $\frac{dW}{dI_{x}}=0$. What you can do is use the fact that $U_{x}$ is monotonically increasing and $U_{y}$ is monotonically decreasing, as functions of $I_{x}$. So the maximum would actually be the point where those two functions cross over in this case. When you're dealing with piecewise functions like this, where the behaviour is different for different values of your parameters, you can't just blanket differentiate - you need to think about the behaviour of your function under different regimes.

On a side note, I'm slightly confused because you say $W=\min{(U_{x},U_{y})}$, but you also say $W=150$. Are these the same quantity?

Tez LaCoyle
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  • I apologise for that, as I said I'm not too sure about a lot of it! Thank you though for your help. I was unsure about the behaviour of the function and what to do with it.

    Also that was a mistake I made; I meant the total income. Sorry!!

     – Kyle Britto Jul 27 '17 at 09:22
  • No need to apologise! Just wanted to warn you before you tried differentiating and then wondering why the derivative is never zero. What really helps a lot with these kind of problems is plotting your function first - knowing what the function looks like gives a huge amount of help in knowing how to find its maxima, etc. – Tez LaCoyle Jul 27 '17 at 09:32