Section 2-4:
Finding Maximums and Minimums
When a quadratic function is
used in finding a maximum or minimum value, you can use the fact that the
vertex happens at x = -b/2a!!
Sample Problem
A rectangular pen is
constructed using one side of the house as a side of the pen. You
have 100 feet of fencing for the other three sides. Find the dimensions
of the greatest size area inside the pen.
Solution!!
Make a diagram!!
A(x) = x(100 - 2x) = 100x -
2x2 = -2x2 + 100x
The maximum happens at x =
-b/2a = -100/(-4) = 25
The dimensions are 25 ft by
25 ft
The greatest area is 25(100
- 50) = 25(50) = 1250 sq ft!!
For a cubic function, we will
use the functions on the Ti-82 to graph the function and then find the
maximum point using the calc function on the calculator to approximate
the answer.
Sample Problem
Squares are cut from the corner
of a rectangular piece of cardboard with dimensions 8 in by 12 in.
The sides will be turned up to form a box with no top. Find the maximum
volume of the box.
Solution: Draw a picture!!
V(x) = x(12 - 2x)(8 - 2x)
Type this function into the
TI-82 calculator!
Using the calc button, estimate
the maximum point on the graph!! You should get ( 1.6, 67.6) rounded
to tenths.
This means that the corner
should be cut to 1.6 inches and the maximum volume will be about 67.6 cubic
inches!!
Sample #3
A manufacturer has 100 tons
of a product that he can sell now for a profit of 5 per ton. For
each week he delays shipment, he can produce an additional 10 tons.
Unfortunately, for each week he delays, the profit decreases 25 cents a
ton. When should he ship to maximize his profit and what is the maximum
profit?
Solution: Let x = number
of weeks to delay
|
Number of tons |
Profit in dollars |
| Now |
100 |
100(5) = $500 |
| In x weeks |
100 + 10x |
5 - .25x |
P(x) = (100 + 10x)(5 - .25x)
= 500 + 25x - 2.5x2 using foil
The maximum value happens at
x = -b/2a = -25/-5 = 5
He should sell in 5 weeks!!
The profit will be 500 + 25(5)
- 2.5(5)2 = $562.50
That's all for this section!!
On to the next!
