Chapter 14, Problem 1RE

To determine

To find: The value of $f\left(5,10\right)$, $f_x\left(x,y\right)$ and $f_y\left(x,y\right)$.

Answer to Problem 1RE

The value of $f\left(5,10\right)=2,900$, $f_x\left(x,y\right)=40$, and $f_y\left(x,y\right)=70$.

Explanation of Solution

Given:

The function is $f\left(x,y\right)=2,000+40x+70y$.

Calculation:

Obtain the value of $f\left(5,10\right)$.

Substitute 5 for x and 10 for and 10 for y in $f\left(x,y\right)=2,000+40x+70y$,

$\begin{array}{c}f\left(5,10\right)=2,000+40\left(5\right)+70\left(10\right)\\ =2000+200+700\\ =2,900\end{array}$

Therefore, the value of $f\left(5,10\right)=2,900$.

Partial derivative of the function $f\left(x,y\right)$ with respect to x (take y as a constant) is computed as follows.

$\begin{array}{c}{f}_x\left(x,y\right)=\frac{\partial }{\partial x}\left(2,000+40x+70y\right)\\ =\frac{\partial }{\partial x}\left(2,000\right)+\frac{\partial }{\partial x}\left(40x\right)+\frac{\partial }{\partial x}\left(70y\right)\\ =0+40\frac{\partial }{\partial x}\left(x\right)+0\\ =40\end{array}$

Thus, the partial derivative of the function $f\left(x,y\right)$ with respect to x is $f_x\left(x,y\right)=40$.

Partial derivative of the function with respect to y (take x as a constant) is computed as follows.

$\begin{array}{c}{f}_y\left(x,y\right)=\frac{\partial }{\partial y}\left(2,000+40x+70y\right)\\ =\frac{\partial }{\partial y}\left(2,000\right)+\frac{\partial }{\partial y}\left(40x\right)+\frac{\partial }{\partial y}\left(70y\right)\\ =0+0+\left(70\right)\frac{\partial }{\partial y}\left(y\right)\\ =70\end{array}$

Thus, the partial derivative of the function $f\left(x,y\right)$ with respect to y $f_y\left(x,y\right)=70$.