The monthly revenue acheived by selling x boxes of candy is calculated to be x(5-0.05x)dollars. The wholesale cost of each box of candy is $1.50. How many boxes must be sold each month to achieve a profit of at least $60.

Can someone please let me know how to set up this equation. I have looked everywhere for help.

You have to define three functions: the revenue function R(x), the cost function C(x) and the profit function P(x)=R(x) - C(x)
We're told R(x)=x(5-0.05x) and C(x)=1.50*x
Thus P(x)=R(x)-C(x)=x(5-0.05x) - 1.50*x
So you want to solve for x when P(x)=60 or
60=x(5-0.05x) - 1.50*x
This is a quadratic equation so be sure to use only positive roots. You might want to graph P(x) to see what it looks like and to see when it starts to fall below 60 because of decreasing marginal profitability.

Net x = the number of boxes sold

Profit = Revenue - Cost = 60

x(5-0.05x) - 1.5 x = 60

3.5 x - .05 x^2 = 60
which can be rewritten
x^2 - 70 x + 1200 = 0

That quadratic equation will have two roots.
Take the lowest positive root.

To solve the quadratic equation x^2 - 70x + 1200 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 1, b = -70, and c = 1200. Substituting these values into the quadratic formula, we can calculate the solutions for x.

x = (-(-70) ± √((-70)^2 - 4(1)(1200))) / (2(1))

Simplifying further:

x = (70 ± √(4900 - 4800)) / 2

x = (70 ± √100) / 2

x = (70 ± 10) / 2

This gives us two possible solutions: x = (70 + 10) / 2 = 80 / 2 = 40, and x = (70 - 10) / 2 = 60 / 2 = 30.

However, since we are looking for the number of boxes that must be sold each month to achieve a profit of at least $60, we need to consider only the positive root. Therefore, the number of boxes that must be sold each month to achieve a profit of at least $60 is 40 boxes.