A company can sell 5000 chocolate bars a month at $0.50 each. If they raise the price to $0.70, sales drop to 4000 bars per month. The company has fixed costs of $1000 per month and $0.25 for manufacturing each bar. What price will maximize the profit?

Let us first arrive at demand function.

Look at the graph. When the price is $0.5, the seller sells 5000 bars. When the price goes up to $0.7, the seller can sell only 400 bars.
From this information, we have to deduce the demand curve.

It is a linear demand curve. use the straight line formula to arrive at demand function.

`p-p_1=(p_2-p_1)/(x_2-x_1)(x-x_1)`

Where -

`x_1=5000`
`p_1=0.5`
`x_2=4000`
`p_2=0.7`

Substitute these values in the above formula

`p-0.5=(0.7-0.5)/(4000-5000)(x-5000)`
`p-0.5=(0.2)/(-1000)(x-5000)`
`p-0.5=-0.0002(x-5000)`
`p-0.5=-0.0002x+1`
`p=-0.0002x+1+0.5`
`p=-0.0002x+1.5` -------------- [Demand Function]

`R=p xx x` ------------------ [Total Revenue ]
`R=(-0.0002x+1.5)xx x`
`R=-0.0002x^2+1.5x`----------------- [Total Revenue Function]

C = Fixed Cost + Variable Cost
C=1000+0.25x -----------------------[Total Cost Function]

`pi= R -C` --------- Profit.
`pi=(-0.0002x^2+1.5x)-(1000+0.25x)`
`pi=-0.0002x^2+1.5x-1000-0.25x`
`pi=-0.0002x^2+1.25x-1000` --------------[Profit function]

`(dpi)/dx=-0.0004x+1.25`
At `(dpi)/dx=0=> -0.0004+1.25=0`
`x=(-1.25)/(-0.0004)=3125`
`(dpi^2)/(dx^2)=-0.0004 < 0`

At `x=3125` second derivative is equatl to zero.

Profit Maximising output `3125`

Substitute `x=3125` in the demand function to find the profit maximising price.

`p=-0.0002x+1.5` -------------- [Demand Function]
`p=-0.0002(3125)+1.5=-0.625+1.5=0.875`

Profit maximising price is `~~$.0.9`