Difference between revisions of "Trivial Inequality"
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After [[Completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function. | After [[Completing the square]], the trivial inequality can be applied to determine the extrema of a quadratic function. | ||
| − | == | + | ==USA AIME 1992, Problem 13== |
| − | < | + | ''Triangle <math>ABC</math> has <math>AB=9</math> and <math>BC: AC=40: 41</math>. What's the largest area that this triangle can have?'' |
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| + | Solution: | ||
| + | First, consider the triangle in a coordinate system with vertices at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>.<br>Applying the distance formula, we see that <math>\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}</math>. | ||
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| + | We want to maximize <math>b</math>, the height, with <math>9</math> being the base. Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>. Recalling that the max/min value of a quadratic is <math>\frac{4ac-b^2}{4a}</math>, we see that <math>\frac{4(-1)(1600) - \left(\frac{3200}{9}\right)^2 }{(-1)(4)} = b^2</math>. Solving for <math>b</math>, <math>b = \frac{40\cdot41}{9}</math> Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>. | ||
| + | |||
| + | ''Solution credit to: 4everwise'' | ||
| + | |||
| + | Note: I am still editing this... | ||
Revision as of 00:47, 18 June 2006
The Inequality
The trivial inequality states that 
for all x. This is a rather useful inequality for proving that certain quantities are non-negative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.
Applications
Maximizing and minimizing quadratic functions
After Completing the square, the trivial inequality can be applied to determine the extrema of a quadratic function.
USA AIME 1992, Problem 13
Triangle 
has 
and 
. What's the largest area that this triangle can have?
Solution:
First, consider the triangle in a coordinate system with vertices at 
, 
, and 
.
Applying the distance formula, we see that 
.
We want to maximize 
, the height, with 
being the base. Simplifying gives 
. Recalling that the max/min value of a quadratic is 
, we see that 
. Solving for , 
Thus, the area is 
.
Solution credit to: 4everwise
Note: I am still editing this...
