Consider the utility function
$$U(x,y) = 6\sqrt{xy}.$$
A consumer has 100 of good $x$ and 2.25 of good $y$. Therefore, his utility is
$$U(100,2.25) = 6\sqrt{100\cdot 2.25} = 6\sqrt{225} = 6\cdot15 = 90.$$
What is his utility approximately if he obtains an extra unit of good $x$ and in return has to give up 0.2 units of good $y$?
According to the approximation of the change of the function value it holds that
$$\Delta U \approx U'_x(x_0,y_0) \Delta x + U'_y(x_0,y_0) \Delta y.$$
It is known that
$$
(x_0,y_0) = (100,2.25), \qquad \Delta x = 1 \qquad \text{and} \qquad \Delta y = -0.2.
$$
We still have to determine the partial derivatives at $(x_0,y_0)=(100,2.25)$:
$$
\begin{align}
U(x,y) &= 6\sqrt{xy} = 6x^{\tfrac{1}{2}}y^{\tfrac{1}{2}}\\[1mm]
U'_x(x,y) &= 6 \cdot \tfrac{1}{2} \cdot x^{-\tfrac{1}{2}}y^{\tfrac{1}{2}} = 3\sqrt{\dfrac{y}{x}}\\
U'_x(100,2.25) &= 3\sqrt{\dfrac{2.25}{100}} = 0.45\\[1mm]
U'_y(x,y) &= 6 \cdot \tfrac{1}{2} \cdot x^{\tfrac{1}{2}}y^{-\tfrac{1}{2}} = 3\sqrt{\dfrac{x}{y}}\\
U'_y(100,2.25) &= 3\sqrt{\dfrac{100}{2.25}} = 20
\end{align}
$$
The change of the function value is approximately
$$\Delta U \approx 0.45 \cdot 1 + 20 \cdot (-0.2) = -3.55.$$
Hence, the function value drops to approximately $90 - 3.55 = 86.45$.
$$U(x,y) = 6\sqrt{xy}.$$
A consumer has 100 of good $x$ and 2.25 of good $y$. Therefore, his utility is
$$U(100,2.25) = 6\sqrt{100\cdot 2.25} = 6\sqrt{225} = 6\cdot15 = 90.$$
What is his utility approximately if he obtains an extra unit of good $x$ and in return has to give up 0.2 units of good $y$?
According to the approximation of the change of the function value it holds that
$$\Delta U \approx U'_x(x_0,y_0) \Delta x + U'_y(x_0,y_0) \Delta y.$$
It is known that
$$
(x_0,y_0) = (100,2.25), \qquad \Delta x = 1 \qquad \text{and} \qquad \Delta y = -0.2.
$$
We still have to determine the partial derivatives at $(x_0,y_0)=(100,2.25)$:
$$
\begin{align}
U(x,y) &= 6\sqrt{xy} = 6x^{\tfrac{1}{2}}y^{\tfrac{1}{2}}\\[1mm]
U'_x(x,y) &= 6 \cdot \tfrac{1}{2} \cdot x^{-\tfrac{1}{2}}y^{\tfrac{1}{2}} = 3\sqrt{\dfrac{y}{x}}\\
U'_x(100,2.25) &= 3\sqrt{\dfrac{2.25}{100}} = 0.45\\[1mm]
U'_y(x,y) &= 6 \cdot \tfrac{1}{2} \cdot x^{\tfrac{1}{2}}y^{-\tfrac{1}{2}} = 3\sqrt{\dfrac{x}{y}}\\
U'_y(100,2.25) &= 3\sqrt{\dfrac{100}{2.25}} = 20
\end{align}
$$
The change of the function value is approximately
$$\Delta U \approx 0.45 \cdot 1 + 20 \cdot (-0.2) = -3.55.$$
Hence, the function value drops to approximately $90 - 3.55 = 86.45$.