Optimization Problems : Make A Calculated Choice

optimization problemsSay you run a business and need to make a decision about how much to spend for marketing, a bigger budget could cause profits to soar. However if your budgets for marketing is much too big, it decreases profit by taking funds away from other expenses. Optimization is a calculus technique that will enable you to solve a problem such as this one.  Be a Calculus 1 Master with this online course that gives you explanations of every calculus topic you will come across in a calculus 1 class.

Optimization problems in computer science and math involve finding the best of all feasible solutions. Problems of optimization have two divisions, depending on whether the variables are discrete or continuous.  Combinatorial optimization problem is another name for an optimization problem with discrete variables. In these types of problems, you need to look for an object such as a graph, permutation or integer from a finite set.  Knowing how to solve optimization problems will allow you to find solutions such as how much materials to utilizes to lessen costs of production as well as what the volume is on a three-dimensional object.  You may also want to take up vectors and this course about calculus introduces you three dimensional coordinate systems covered in Calculus 3.

The best way to find the minima or the maxima of a function (the points where functions are at their smallest or biggest) is through optimization. Optimization can be constrained, where your subject has a constraint. It can also be unconstrained, where there are no constraints to consider. Usually, constraints are conditions you can describe using another function of math.

In problems of optimization, we are looking for the smallest or the largest value that a function can take.  When it comes to optimization problems in the real world, we will be looking at a functions’ smallest or largest value when it is under constraints. As a matter of fact, you might be interested in this course called Calculus 2- Application of Integrals that shows you how to apply calculus for solving real world problems.  Usually, a constraint is a kind of circumstance that the equation describes that needs to be positively, absolutely true no matter what solution you come up with. Once you read the problem carefully, the first thing you should do is get all the constraints identified. By the way, anytime you see a fixed value indicated in a problem, this is the constraint.  The moment you have made an identification of the amount that needs to be optimized, the rest is fairly simple to solve. Here is an article called Calculus Derivatives that tells you exactly how to solve these.

Optimization Problem #1

You need to build a box with a base width that is one third of its base length. To build the top and bottom, the material to be used cost ten dollars per square foot and to build the sides, it costs six dollars per square foot. If the box needs to have a fifty cubic foot volume, find the dimensions that will be the least costly to get the box built.


h /                                                    /

/——————————-/*  w


To solve the problem, create a corresponding model using math:

Constraint: 50 = 3w2h = lwh

Minimize: 48wh + 60w2= 6(2wh + 2 * 3w * h) + 10 (2 * 3w * w) = 6 ( 2wh + 2lh) + 10 (2 lw) = C


Constraint:  50 = 3w2h à 50/3w2 à C = 60w2 + 800/w = 48w * 50/3w2= 60w2 + 48wh

To solve for this, use the first derivative test of absolute extreme values and so when you substitute all the values the minimum cost is C = $637.60


You need to get a field fenced off. You have a building on one of the field’s sides that won’t need a fence and you have 500 feet of material you can use for fencing. Find the field’s dimensions to enclose the maximum area.

Solution: There are 2 functions in this problem. The first is what you are trying to get optimized and the next are the constraints. Sketch the situation as this frequently helps you arrive at the right equation:



\                                       \  y



You know you have 500 feet of fence material so the optimized function is the area and the constraint is the fencing material:

Constraint: 500 = x + 2y

Maximize: A = xy

So you know how to find the smallest or the largest function value as long as there is just one variable. The constraint and the area have 2 variables so you know that you need to solve for constraint x:

500 – 2y = x

Substitute this into the function for Area: 500y – 2y2 = (500 – 2y) y = A (y)

Find the biggest value this has on the [0, 250] interval as this corresponds to taking 0=y or that the fence has no sides and 250 = y, since there are just 2 sides and to use the whole 500 feet then each side must be this size.

Keep in mind that the interval’s endpoints will not make sense if you want some area enclosed since this would both give an area of zero. However, on ‘y’ there is a set of limits which tell you that if you have a least value somewhere between both endpoints will mean that you can use a process to find the area’s maximum value.

Basically, what we’ve got here is a continuous function’s maximum value on an interval that is closed which will occur at end or critical points. To get the critical points:

500 – 4y = A (y)

You get a lone critical point of 125 = y when setting this to be equal to ‘0’ If you plug this into the equation you get A = 31,250 feet.  So when this method is used, then this must be the largest area possible, since zero is the area at either end point.

To get the dimensions, let’s solve for ‘x.’

500 – 2 (125) = 250 = x

So the field’s dimension that gives the broadest area limited by the fact that you only have fencing material of five hundred feet is 125 by 250.

To use this method, there needs to be finite endpoints in the possible optimal values. Also, the optimized function needs to be continuous on the endpoints. If you meet these conditions, you know that the optimal values, whether the minimum or maximum will happen at either of the range’s endpoints or within the range at a critical point.

Optimization Problem #2

A man goes from a river bank point A on a straight river three kilometers wide and wants to reach the opposite bank, eight kilometers on the opposite bank point B as fast as he can. There are three ways he can proceed:

  1. Row to some point between B and C on point D and then run to B
  2. Row to B directly
  3. Row his boat across the river directly on another point C before running over to point B.

If he is able to run 8 km/h and row 6 km/h where should he land to reach B the fastest?

The solution for this problem is to let the distance be ‘x’ from point C to D and the running distance is /DB? =8-x. The rowing distance is given by the Pythagorean Theorem as /AD/ = square root of x2 + 9. Use the equation: time=distance/rate, substitute all the values in to it to come up with the solution.

You will find that the smallest value of time occurs when x equals 9 over square root of seven, which is the minimum value of time. Thus the man ought to land his boat about 3.4 kilometers down the stream from his point of origin.

Hope this helps! Here is a course that shows you more essentials to a calculus introductory course called Calculus 1 Essentials.