Use quadratic functions. Suppose that the equation p(x)=−2x2+280x−1000, where x represents the number of items...

Step 1
We have,
1) ${\displaystyle p\left(x\right)=-2x^2+280x-1000}$
This is a quadratic function with ${\displaystyle a=-2,b=280}$ and ${\displaystyle c=-1000\times =-\frac{b}{2a}}$
Since ${\displaystyle a=-2<0}$, which create open downward parabola because 'a' is negative which therefore, creates a maximum at the vertex.
Let us determine the number of terms that should be produced to maximize the cost by find the x-value of the vertex.
We know that, vertex of parabola be:
2) ${\displaystyle x=-\frac{b}{2a}}$
Substitute the value of ${\displaystyle a=-2}$ and ${\displaystyle b=280}$ in equation (2), we get
${\displaystyle \Rightarrow x=-\frac{280}{2\times (-2)}}$
${\displaystyle \Rightarrow x=-\frac{280}{(-4)}}$
${\displaystyle \Rightarrow x=\frac{280}{4}}$
${\displaystyle \Rightarrow x=70}$
Therefore, the value of x-coordinates is 70
Hence, the number of items that maximize the profit is 70 items.