Example 4

Consider the function $U(x,y) =2x^{\tfrac{1}{3}}y^{\tfrac{2}{3}}$. Determine the tangent line to the indifference curve of $U(x,y)$ at $(x,y)=(1,1)$.

The indifference curve at $(x,y)=(1,1)$ has the value
$$U(1,1) = 2\cdot1^{\tfrac{1}{3}}\cdot 1^{\tfrac{2}{3}} = 2.$$
If we rewrite $U(x,y)=2$ to $y(x)$, then we get (see Example 2):
$$y(x) = \dfrac{1}{\sqrt{x}} = x^{-\tfrac{1}{2}}.$$
We have to determine the tangent line to the graph of the function $y(x)$ in the point $x=1$. The general form of a tangent line is
$$t(x) = ax + b,$$
where $a$ is the slope of the tangent line and $b$ is the point of intersection with the $y$-axis. If $t(x)$ is the tangent line of the graph of $y(x)$ in the point $x=1$, then it has to hold that:
$$(1)~y(1) = t(1) \qquad \text{and} \qquad (2)~y'(1) = t'(1).$$
We start with $(2)$:
$$\begin{align} y'(x) &= -\dfrac{1}{2} x^{-\tfrac{1}{2}-1} = \dfrac{-1}{2x\sqrt{x}}\\ y'(1) &= \dfrac{-1}{2\cdot 1\cdot \sqrt{1}} = -\dfrac{1}{2}\\[1mm] t'(x) &= a\\ t'(1) &= a. \end{align}$$
In other words: $a=-\tfrac{1}{2}$, or $t(x) = -\tfrac{1}{2}x + b$. We now use $(1)$ to determine $b$:
$$\begin{align} y(1) &= \dfrac{1}{\sqrt{1}} = 1\\ t(1) &= -\tfrac{1}{2} \cdot 1 + b = -\tfrac{1}{2} + b \end{align}$$
Finally, it holds that
$$1 = -\tfrac{1}{2} + b, \qquad \text{hence} \qquad b = 1 + \tfrac{1}{2} = \tfrac{3}{2}.$$
The tangent line to the level curve of $U(x,y) = 2x^{\tfrac{1}{3}}y^{\tfrac{2}{3}}$ at $(1,1)$ is therefore equal to:
$$t(x) = -\tfrac{1}{2}x + \tfrac{3}{2}.$$