Optimization

=__**Optimization**__= The graph of a function always has a highest range value called the absolute maximum and a lowest value called the absolute minimum. This value can potentially be found at multiple domain values or even an interval. There are two situations where an optimal value must be found, open and closed intervals.

An open interval: find all values at which the derivative of a function is either increasing then decreasing or decreasing then increasing. The values between these indicate either a maximum or minimum, but must be individually tested in order to obtain a certifiable value, the highest or lowest being the maximum and minimum. It i a maximum if increasing the decreasing, or a minimum if decreasing then increasing.

A closed interval: in addition to the steps described in an open interval, the endpoints must also be tested.

Optimization problems use the concept of finding the highest output at a given input in order to find the maximum value for a moddled function, usually with multiple variables. The trick is to narrow down the number of variables to one using constraints given in the original explanation by means of substitution, then taking the derivative of the resulting function to find a singular value at which the desired output is maximum, then using the formula to solve for the other variable.

Example: Zach's transition lenses' percent absorbance is moddled by y=3a 3 +b 3 +8c 2 - 2d where a is the concentration of silver, b is the concentration of chloride, c is the temperature in degrees celcius after solar heating, and d is the deviation in thickness due to thermal expansion/contraction. Due to fundamental ionic properties, the concentration of silver ion and chloride ion must be equal. The deviation in the thickness of glass as a function of temperature is moddled by the function d=9c/2. Since the same amount of atoms remain and the volume changes, the ionic concentration of silver is augmented by the function a=81d 3. Find the temperature at which Zach's glasses are most efficent at absorbing ultraviolet rays knowing that temperature cannot fall below -273 degrees Celsius or go above 1000 degrees Celsius without melting the glass. Percent absorbance cannot be negative.

Answer: We know y=3a 3 +b 3 +8c 2 -2d a=b d=9c/2 so c=2d/9 a=d 3 /3 Start substituting y=4a^3+8c^2-2d y=4d^9/81+8c^2-2d y=4d^9/81+32d^2/81-2d Take the derivative using the shortcut y'=4d^8/9+64d/81-2 The zeroes are -1.26976 and 1.12168 by use of a calculator Because a maximum is desired, we must find the point where the function is increasing before and decreasing afterwards: y' is positive then negative. y(-273)=-4.159(10^20) y'(-2)=110.198 y(-1.26976)=2.75276 y'(0)=-2 y(1.12168)=-1.6075 y'(2)=113.358 y(1000)=4.9383(10^25) So 1000 degrees celcius would be ideal