You should use the equation that relates profit, cost and revenue such that:

`P(x) = R(x) - C(x)`

`P(x)` represents the profit function

`R(x)` represents the revenue function

`C(x)` represents the cost function

Substituting `6x - 1/100 x^2` for `R(x)` and `50 + 3x` for `C(x)` yields:

`P(x) = 6x - 1/100 x^2 - 50 - 3x`

You need to combine the terms that contain the same powers of x such that:

`P(x) = - 1/100 x^2 + 3x - 50`

You need to notice that the equation of provit function is a quadratic equation and since the leading coefficient is negative, the maximum value of the profit function `y = (4ac - b^2)/(4a)` is reached at `x = -b/(2a).`

Identifying the coefficients a,b,c yields:

`a = -1/100, b = 3, c = -50`

`x = -3/(-2/100) => x = 300/2 => x = 150`

`y = (200/100 - 9)/(-4/100) => y = 700/4 => y = 175`

**Hence, evaluating the number of products produced per week to reach the maximum profit yields that for `150` products produced per week, the profit is of `$175.` **

The formula of profit is:

Profit = Revenue - Cost

Plug-in Revenue= R(x) and Cost=C(x).

Profit= R(x) - C(x)

Since `R(x) = 6x-1/100x^2` and `C(x)=50+3x` , then,

Profit = `6x-1/100x^2 - (50+3x)`

Profit = `6x-1/100x^2 - 50-3x`

Profit =`-1/100x^2+3x-50`

Notice that the profit function is in quadratic form. So its graph is a parabola.

In parabola, if the coefficient of x^2 is negative, the vertex is the maximum point.

Since the coefficient of x^2 in the profit function is negative, then its vertex is the maximum point.

To determine its vertex (h, k), use the formula,

`h=-b/(2a)`

where a is the coefficient of x^2 and b is the coefficient of x.

`h = -3/(2*(-1/100)) = -3/(-1/50) = -3*-50/1=150`

Then, substitute x with the value of h to the profit function to get k.

`k=Profit = -1/100x^2+3x-50=-1/100*150^2 +3*150-50`

`k=-225 +450-50=175`

Base on this, the vertex is (150, 175).

**Hence, the company has to sell 150 products per week to attain the maximum profit of $175 per week.**