Answer:
The profit is maximum when 28 units are sold.
Step-by-step explanation:
The given revenue function is
[tex]R(x)=30x-0.5x^2[/tex]
The given cost function is
[tex]C(x)=2x+8[/tex]
Profit is the difference of revenue and cost. So the profit function is
[tex]P(x)=R(x)-C(x)[/tex]
[tex]P(x)=(30x-0.5x^2)-(2x+8)[/tex]
[tex]P(x)=30x-0.5x^2-2x-8[/tex]
On combining like terms, we get
[tex]P(x)=-0.5x^2+28x-8[/tex]
It is a quadratic function. Leading coefficient is negative so it is a downward parabola. Vertex of a downward parabola is the point of maxima.
If a quadratic function is [tex]y=ax^2+bx+c[/tex], then the vertex of the parabola is
[tex](-\frac{b}{2a},f(-\frac{b}{2a}))[/tex]
In the profit function a=-0.5, b=28 and c=-8, the function is maximum at
[tex]x=-\frac{b}{2a}[/tex]
[tex]x=-\frac{28}{2(-0.5)}[/tex]
[tex]x=28[/tex]
Therefore the profit is maximum when 28 units are sold.
The number of units that yield the maximum profit is 56
The cost and revenue functions are given as:
C(x) = 2x + 8
R(x) = 30x - 0.5x^2
Calculate the profit function
P(x) = R(x)- C(x)
This gives
P(x) = 30x - 0.5x^2 - 2x - 8
Evaluate the like terms
P(x) = -0.5x^2 + 28x -8
Differentiate
P'(x) = -0.5x + 28
Set to 0
-0.5x + 28 = 0
Collect like terms
-0.5x =- 28
Divide both sides by -0.5
x =56
Hence, the number of units that yield the maximum profit is 56
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