Difference between revisions of "Optimization"
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If <math>a>0</math>, then the quadratic <math>ax^2+bx+c=0</math> reaches its minimum when <math>x=-\frac{b}{2a}</math>, while when <math>a<0</math>, the quadratic reaches its ''maximum'' when <math>x=-\frac{b}{2a}</math>. | If <math>a>0</math>, then the quadratic <math>ax^2+bx+c=0</math> reaches its minimum when <math>x=-\frac{b}{2a}</math>, while when <math>a<0</math>, the quadratic reaches its ''maximum'' when <math>x=-\frac{b}{2a}</math>. | ||
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| + | ===Alternative=== | ||
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| + | The alternative requires [[calculus]]. The optimum point has a [[tangent]] line with a slope of 0. Thus, calculating the [[derivative]] and setting it equal to 0 will also give an answer. Assuming that the original [[quadratic]] is <math>ax^2+bx+c</math>, the derivative is <math>2ax+b</math>. Setting <math>2ax+b = 0</math>, we find that <math>x = -\frac{b}{2a}</math> . | ||
Revision as of 15:48, 27 November 2016
The optimization of a quadratic equation is the process to find the maximum or minimum of said quadratic.
Process
It involves converting a quadratic to the standard form 
by completing the square. Then by the Trivial Inequality, the maximum or minimum (it depends on which way the graph of the quadratic is facing) is 
.
"Formula"
To optimize a quadratic, one might use the method described above, or one could use this other, smoother, method:
If 
, then the quadratic 
reaches its minimum when 
, while when 
, the quadratic reaches its maximum when .
Alternative
The alternative requires calculus. The optimum point has a tangent line with a slope of 0. Thus, calculating the derivative and setting it equal to 0 will also give an answer. Assuming that the original quadratic is 
, the derivative is 
. Setting 
, we find that 
.
