Section 24:
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 
2x^{2} = 2x^{2} + 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 Ti82 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
TI82 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  2.5x^{2} 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!