optimization: Profit = (x^3) - 48x^2 + (576x) - 1500

[skyblue]

Profit= (x^3) - 48x^2 + (576x) - 1500

x= the number of units sold.
How many units will be sold to maximize the company's profit?

 

[soroban]

Re: optimization

Hellom skyblue!

You've never done a max/min problem before?

Profit: \(\displaystyle \,P(x)\;= \:x^3\,-\,48x^2\,+\,576x\,-\,1500\)
. . \(\displaystyle x\) = the number of units sold

How many units will be sold to maximize the company's profit?

Set the derivative equal to 0 and solve . . .

\(\displaystyle P'(x)\:=\:3x^2\,-\,96x\,+\,576\:=\:0\)

Divide by 3: \(\displaystyle \:x^2\,-\,32x\,+\,192\:=\:0\)

Factor: \(\displaystyle \,(x\,-\,8)(x\,-\,24)\:=\:0\)

. . and we have two roots: \(\displaystyle \:x\:=\:8,\:24\)


Test these with the second derivative: \(\displaystyle \,f''(x)\:=\:6x\,-\,96\)

. . \(\displaystyle f''(8)\,=\,6(8)\,-\,96\:=\:-48\;\) negative, concave down, \(\displaystyle \cap\), maximum

. . \(\displaystyle f''(24)\,=\,6(24)\,-\,96\:=\:+48\;\) positive, concave up, \(\displaystyle \cup\). minimum.


Therefore, for maximum profit: \(\displaystyle x\,=\,8\)