Function optimization . There is a wide variety of problems and applications that have the following purposes: find the minimum area, the lowest cost, the optimal form, the least resistance, the maximum benefit, the greatest scope … All these problems are included within the Function Optimization category and can be solved by applying the differential calculation .
Summary
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- 1 Optimization problems
- 1 Steps for solving optimization problems
- 2 Source
- 3 See also
Optimization problems
Many of the problems that arise in practice on a daily basis are related, in one way or another, to finding the maximum and minimum values of a function , and even more, determining for which values of the independent variable these are reached. These problems are generally called optimization problems.
In general terms, an optimization problem consists of finding the minimum value or minimizing, or finding the maximum value or maximizing, a certain function, in such a way that they satisfy certain given conditions.
The optimal solution or solutions are those for which the constraints of the problem are satisfied and the value of the function is minimum or maximum.
Steps for solving optimization problems
- Draw an analysis figure (if necessary)
- The function that must be maximized or minimized is proposed.
- An equationis proposed that relates the different variables of the problem, in the event that there is more than one variable.
- A variable is removed from the equation and replaced in the function so that we have only one variable.
- The function is derivedand equals zero, to find the local extremes .
- The 2nd derivative is performed to check the result obtained.
- Answer the problem question.
Example 1:
You need a rectangular surface fenced on three sides with wire mesh and on the fourth side with a stone wall. 20 linear meters of wire mesh are available. Calculate the dimensions that the surface must have so that its area is as large as possible.
Resolution
The figure shows the surface.
Its area is A = xy with x, and the sides of the rectangle .
Since 20 meters of wire mesh are available, then x + y + x = 20, that is, y = 20-2x
A = xy = x (20-2x) = 20x-2x 2 with 0 <= x <= 10
We find the maximum of function A.
A´ (x) = 20-4x
20-4x = 0
x = 5
A´´ (x) = – 4 for all x, so at x = 5 there is a maximum. We compare this maximum with the values of the ends of the interval:
A (0) = 0, A (10) = 0
Then the maximum is absolute and is the value sought
The value of y = 20-2 (5) = 10
Rta. The side enclosed by the wall should measure 10m and the other side 5m.
Example 2:
Example 3:
The total cost of production of x units of a product is C T = 5 / 2 x 2 + 20x weights and unit price is 50- p = x / 2 weights. Find the number of units that must be sold for maximum profit.
C T = 5 / 2 x 2 + 20x
p = 50x / 2 I (x) = x. (50x / 2 ) = 50x-x 2 / 2
G (x) = I (x) -C T
G (x) = 50x-x 2 / 2 – ( 5 / 2 x 2 + 20x)
G (x) = 30x-3x 2
G´ (x) = 30-6x
30-6x = 0
x = 5 Rta. The number of units to be sold is 5.
