Answer:
One needs to sell 220 units of the product to earn a maximum profit of $4,870.
Step-by-step explanation:
The revenue is [tex]R(x)=50x-0.1x^2[/tex].
The cost is [tex]C(x)=6x+30[/tex]
The profit is given by,
Profit = Revenue - Cost
[tex]P(x)=50x-0.1x^2-(6x+30)[/tex]
[tex]P(x)=-0.1x^2+44x+30[/tex] ....(1)
To maximize the above profit function in (1) we find its critical points.
[tex]P'(x)=-0.2x+44[/tex]
[tex]-0.2x+44=0[/tex]
[tex]x=\frac{44}{0.2}[/tex]
[tex]x=220[/tex]
To verify this critical point is a point of maximum, we calculate the second derivative of P(x) to get
[tex]P''(x)=-0.2<0[/tex]
Hence from the second derivative test, x = 220 is a point of maximum for P(x).
Substitute x in profit,
[tex]P(220)=-0.1(220)^2+44(220)+30[/tex]
[tex]P(220)=4870[/tex]
One needs to sell 220 units of the product to earn a maximum profit of $4,870.
