Difference between revisions of "022 Exam 2 Sample A, Problem 8"

Use differentials to approximate the change in profit given ${\displaystyle x=10}$  units and ${\displaystyle dx=0.2}$  units, where profit is given by ${\displaystyle P(x)=-4x^{2}+90x-128}$.

Foundations:
A differential is a method of linearly approximating the change of a function. We use the derivative of the function at an initial point ${\displaystyle x_{0}}$ as the slope of a line, and use the standard relation
${\displaystyle m\,=\,{\frac {\Delta y}{\Delta x}},}$
where ${\displaystyle \Delta y}$ represents the change in ${\displaystyle y}$ values, and ${\displaystyle \Delta x}$ represents the change in ${\displaystyle x}$ values. Due to the use of the derivative ${\displaystyle f'\left(x_{0}\right)}$ as the slope, we usually rewrite this using ${\displaystyle dy}$ and ${\displaystyle dx}$ to indicate the relative changes. Thus,
${\displaystyle f'(x_{0})\,=\,m\,=\,{\frac {dy}{dx}}.}$
We can then rearrange this to find ${\displaystyle dy=f'(x_{0})\cdot dx.}$

Solution:

Step 1:
First, we must find the derivative. We have ${\displaystyle P'(x)=-8x+90}$.
Step 2:
We need the derivative at our initial point, or ${\displaystyle x_{0}=10}$. This is
${\displaystyle P'(x_{0})\,=\,P'(10)\,=\,-8(10)+90\,=\,10.}$
Step 3:
Finally, we plug in the values to find
${\displaystyle dy\,=\,P'(x_{0})\cdot dx\,=\,10\cdot 0.2\,=\,2.}$
Note that if a teacher gives you units (thousands of dollars, dollars, cubits...), you should include them in your answer.
${\displaystyle dy\,=\,2.}$