Answer:
[tex](a)\ q=3.5[/tex]
[tex](b)\ t_{max} = 120.25[/tex]
(c) No maximum
Explanation:
Given
[tex]p = 108 - 5q[/tex]
[tex]c = -12q + q^2[/tex]
Solving (a): The price at maximum profit.
The profit (t) is calculated using the following function
[tex]t = p - c[/tex] --- price- cost
So, we have:
[tex]t = 108 - 5q - (-12q + q^2)[/tex]
Open bracket
[tex]t = 108 - 5q +12q - q^2[/tex]
[tex]t = 108 +7q - q^2[/tex]
Rewrite as:
[tex]t = - q^2 + 7q + 108[/tex] --- this is the profit function
A quadratic function is represented as:
[tex]y = ax^2 + bx + c[/tex]
The maximum is:
[tex]x = -\frac{b}{2a}[/tex]
By comparison:
[tex]q = -\frac{7}{2*-1}[/tex]
[tex]q=3.5[/tex] ----- price at maximum profit
Solving (b): The maximum profit
The profit function is:
[tex]t = - q^2 + 7q + 108[/tex]
The maximum is:
[tex]t_{max} = -3.5^2 + 7 * 3.5 +108[/tex]
[tex]t_{max} = 120.25[/tex]
Solving (c): Maximum cost
We have:
[tex]c = -12q + q^2[/tex]
The maximum price of the above function is:
[tex]q = -\frac{-12}{2*1}[/tex] ---- from x = -b/2a
[tex]q = \frac{12}{2}[/tex]
[tex]q = 6[/tex]
So, the maximum cost is:
[tex]c_{max} = -12 * 6 + 6^2[/tex]
[tex]c_{max} = -36[/tex]
The cost function has no maximum; only minimum
