Example: Finding the sales level for maximum profit given the equation for a profit function. If the sales level is fractional, round up & down to check the value of the profit at both values. Normally, in business, we do not sell a fraction of an item. Given: P(x) = -x³ +77x² -150x+200 where x is between 0 and 70.

Method 1: Using your calculator. First, code in the profit function using the y= menu. Make sure that the first negative sign is from the bottom of your calculator.

Make sure your table is set up properly. Go to the 2^nd window menu. For the independent variable it should be set at ASK & for the dependent variable it should be set at Auto. Keep this setting though out the entire problem.

Before setting a viewing window, get some idea of the values involved by going to 2^nd graph. A table for the x & y values should be displayed. The y values are the P(x) values for each x input. Since x is restricted between 0 & 70, input values starting at 0 in intervals up to 70 and take a note of the magnitude of the numbers involved. Do not skip too many since you could miss the maximum value.

You will notice that the values for P(x) will range from 200 to over 60000 then start to decrease. So, for the x window, use a low value (min X) of 0 and a high value (max x) of 70. For the y window, use a min of 0 and a max of 80000 (to be safe).

Press graph to check your graph. The maximum (highest point) should be visible. If not, adjust your window (y range) until it is clearly visible. Decreasing the range on y will “stretch” your graph up/down. Increasing the y range will compress your graph vertically. Decreasing the range on x will “stretch” your graph right/left. Increasing the range on x will compress it horizontally.

Press 2nd Calc, then go to menu 4, enter. The graph should be visible. Press trace. A marker on the curve should be visible & can be moved along the curve using the right & left arrows. The calculator will prompt you for a left guess for x. Enter any value for x on the left side of the highest point, enter. Then do the same for any value for x on the right side, enter. The calculator will then prompt you for a guess. Enter any x value between your left & right values. Then the answer for x at the maximum will be displayed at the bottom along with its y value.

If all procedures are followed correctly, it should be x=50.34. Check x=51 using the table, that gives a profit of $60,176. At x=50, the profit is $60,200. So, x=50 produces the maximum profit of $60,200.

The value for the profit at the right end point, x=70, is $24,000. So, that value for x can be eliminated. So can x=0.

Method 2: The derivative method. Find the values for x for which P'(x)=0. Since this is a quadratic equation (does not factor), the quadratic formula from algebra must be used. Take the positive value for x. Solve: -3x²+154x-150=0 or 3x²-154x+150=0, by multiplying both sides by a negative 1.
x = -(-154)+sqr[(-154)²-4(3)(150)] all divided by 2(3) = 50.34

Note: If the profit function is 4^th degree or higher, the calculator method must be used.
So, know it well.

Note: A sketch of the situation and should help you understand the problem better. In most cases, a sketch is not required (only if asked). You will notice that the function increases to the value of its maximum point where x=50.34. The tangent line drawn will have a slope of zero (horizontal tangent).