A company sells X units of a product whose profits function is given by the quadratic model P(X) =60X-X^2 = X(60-X)

A) How many units should the company sell in order to maximize their profits

B) What will be the maximum profit if the # of units found in part (A) were sold

C) what is the maximum number of units the business can sell and still make a pofit

2. Originally Posted by waterboy
A company sells X units of a product whose profits function is given by the quadratic model P(X) =60X-X^2 = X(60-X)

A) How many units should the company sell in order to maximize their profits

B) What will be the maximum profit if the # of units found in part (A) were sold

C) what is the maximum number of units the business can sell and still make a pofit

Using $P(x)=ax^2+bx+c$, the maximum point is reached when $x=\frac{-b}{2a}$

So, in $P(x)=-x^2+60x \;,x=\frac{-60}{-2}=30$.

Profit would be $P(30)=900$.

3. Thank you masters but can you show me how you rearranged THE PROBLEM

60X-X^2=X(60-X)

i DID THIS

60X-X^2=60X-X
X^2=X^2

60X=X^2+60X-X

where did you get a,b and c

4. Originally Posted by waterboy
Thank you masters but can you show me how you rearranged THE PROBLEM

60X-X^2=X(60-X)

i DID THIS

60X-X^2=60X-X
X^2=X^2

60X=X^2+60X-X

where did you get a,b and c
The standard form of a quadratic function is:

$f(x)=ax^2+bx+c=0$

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

Your function was $P(x)=60x-x^2$

If you rearrange your function into standard form, it would look like this:

$P(x)=-x^2+60x$, where a = -1, b = 60.

This is the graph of a parabola that opens downward. The vertex would be the maximum point on the parabola. The x-coordinate of the vertex is given by the formula:

$x=\frac{-b}{2a}$

This formula is derived by finding the x-intercepts and dividing by 2, but that's another story. Just remember this formula.

Substituting the coefficients into that formula, I found the x-coordinate to be 30.

Substituting P(30) into $P(x)=-x^2+60x$, I found the y-coordinate of the vertex, which is the maximum value (profit) when 30 units are sold. P(30) = 900.