Answer :
To find the minimum and maximum selling prices at which the company can make a profit, we need to determine when the profit function is greater than zero.
Given:
- Revenue function: [tex]\( R(x) = 1248x - 8.32x^2 \)[/tex]
- Cost function: [tex]\( C(x) = 36400 - 83.2x \)[/tex]
We define the profit function as:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
[tex]\[ P(x) = (1248x - 8.32x^2) - (36400 - 83.2x) \][/tex]
[tex]\[ P(x) = 1248x - 8.32x^2 - 36400 + 83.2x \][/tex]
[tex]\[ P(x) = (1248 + 83.2)x - 8.32x^2 - 36400 \][/tex]
[tex]\[ P(x) = 1331.2x - 8.32x^2 - 36400 \][/tex]
To find the critical points where the profit is zero, we solve:
[tex]\[ P(x) = 0 \][/tex]
[tex]\[ 1331.2x - 8.32x^2 - 36400 = 0 \][/tex]
The real and non-negative solutions to this quadratic equation are:
[tex]\[ x = 35.0000000000000 \][/tex]
[tex]\[ x = 125.000000000000 \][/tex]
So, for the company to make a profit, the selling price can go no lower than \[tex]$35.00 and no higher than \$[/tex]125.00.
Therefore:
- The selling price can go no lower than \[tex]$35.00 and - no higher than \$[/tex]125.00
Given:
- Revenue function: [tex]\( R(x) = 1248x - 8.32x^2 \)[/tex]
- Cost function: [tex]\( C(x) = 36400 - 83.2x \)[/tex]
We define the profit function as:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
[tex]\[ P(x) = (1248x - 8.32x^2) - (36400 - 83.2x) \][/tex]
[tex]\[ P(x) = 1248x - 8.32x^2 - 36400 + 83.2x \][/tex]
[tex]\[ P(x) = (1248 + 83.2)x - 8.32x^2 - 36400 \][/tex]
[tex]\[ P(x) = 1331.2x - 8.32x^2 - 36400 \][/tex]
To find the critical points where the profit is zero, we solve:
[tex]\[ P(x) = 0 \][/tex]
[tex]\[ 1331.2x - 8.32x^2 - 36400 = 0 \][/tex]
The real and non-negative solutions to this quadratic equation are:
[tex]\[ x = 35.0000000000000 \][/tex]
[tex]\[ x = 125.000000000000 \][/tex]
So, for the company to make a profit, the selling price can go no lower than \[tex]$35.00 and no higher than \$[/tex]125.00.
Therefore:
- The selling price can go no lower than \[tex]$35.00 and - no higher than \$[/tex]125.00