Let's solve the problem step by step.
We are given the revenue function:
[tex]\[ R(x) = -x^2 + 50x + 300 \][/tex]
This is a quadratic function of the form [tex]\( R(x) = ax^2 + bx + c \)[/tex], where:
- [tex]\( a = -1 \)[/tex]
- [tex]\( b = 50 \)[/tex]
- [tex]\( c = 300 \)[/tex]
Quadratic functions form a parabola, and since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a \)[/tex]) is negative, the parabola opens downwards. This means the maximum revenue occurs at the vertex of the parabola.
The x-coordinate of the vertex of a parabola given by [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plug in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = -\frac{50}{2(-1)} \][/tex]
[tex]\[ x = -\frac{50}{-2} \][/tex]
[tex]\[ x = 25 \][/tex]
So, 25 boxes of paper should be sold to maximize revenue.
Next, we calculate the maximum revenue by substituting [tex]\( x = 25 \)[/tex] back into the revenue function [tex]\( R(x) \)[/tex]:
[tex]\[ R(25) = - (25)^2 + 50(25) + 300 \][/tex]
[tex]\[ R(25) = -625 + 1250 + 300 \][/tex]
[tex]\[ R(25) = 925 \][/tex]
Therefore, the maximum revenue is [tex]$925.
To summarize:
- The number of boxes of paper that should be sold to maximize revenue is 25.
- The maximum revenue is $[/tex]925.
