Question

Without using calculus or any other advanced math, the MS Solver plug-in can be used to find the input value for x that results in a maximum value for a function f(x). The price x is in the Solver "variable cell" and the function 1000x – 300x^2 is the Solver "objective." Question: What is the price x that maximizes weekly revenues?

Answer 1

Answer:

The weekly revenue is maximum at x=1.67.

Explanation:

The given function is

$f(x)=1000x -300x^2$         .... (1)

where, f(x) is the total revenue at price x.

We need to find the price x at which the weekly revenue is maximum.

The leading coefficient of the given function is -300, which is a negative number. So, it is a downward parabola and vertex of a downward parabola is the the point of maxima.

If a parabola is defined as

$g(x)=ax^2+bx+c$           ... (2)

then the vertex of the function is

$(-\frac{b}{2a},g(-\frac{b}{2a}))$

From (1) and (2) it is clear that

$a=-300,b=1000, c=0$

The given function is maximum at

$-\frac{b}{2a}=-\frac{1000}{2(-300)}$

$-\frac{b}{2a}=\frac{10}{6}$

$-\frac{b}{2a}=1.66667$

$-\frac{b}{2a}\approx 1.67$

Therefore the weekly revenue is maximum at x=1.67.