Subject: Calculus

# Maxima And Minima

## Who is this Maxima and Minima?

Maxima and minima are not humans nor tv or cartoon characters. Maxima and minima refers to the maximum and minimum values of a function. Using first derivative test and second derivative test, we were able to find several function properties such as concavity, intervals of concavity relative maximum and minimum values.

## Okay so how do we solve for maxima and minima?

Now, when we represent functions as quantities and we seek to find its largest values and smallest values possible, then it is called as problems involving maxima and minima. In this problem, we shall deal with functions as quantities and we'll try to solve the largest or smallest possible values of that function. The following are the steps in solving the maxima and minima of a certain quantity.

1. Sketch and draw the figure whenever possible. Some minima and maxima problems involves area, regions, volumes and shapes. Thus, it is very best to draw the figure to have a better view and understanding of the problem.

2. Label the quantities in the figure above. Make sure that all the quantities mentioned in the problem is depicted and labeled properly.

3. Express the quantity which is being maximized or minimized as a function of the other known quantities. In this way, it would be easy to differentiate the function directly.

4. Find the critical points of the function obtained in step three by doing the first derivative method. This can be done by differentiating the function obtained in step three and equate to zero and then solve for the critical points.

5. Differentiate the function twice and solve the value of the second derivative at the critical points. When that value is negative, then it corresponds to the maximum value. If it is negative, then is the minimum value, otherwise, the function has no maximum or minimum value.

### Illustration

Now, let's apply the above steps to solve maxima and minima problems. In our first problem, we want to make an open-top rectangular box and we want to maximize the volume of the box under a tight budget of $2400. Also, the material for the base of the costs$8 per square foot and the sides cost $2. What would be our box's dimension such that we get the most out of our budget? #### Step 1: Let us draw the configuration of the problem. #### Step 2: Labeling of quantities. As for the rectangular box depicted in the figure above, we assume that the base is a square with side length s and the height is h. #### Step 3: Function expression We know that the volume of our box can be expressed as V=s*s*h=s^2h. Since we want the volume of the box to be maximized, then our function would be the expression of the volume V which is a function of the sides s and the height h. Note that V, h and s are all unknown, thus, we can't solve this problem if that's the case. Therefore, we must relate atleast one of the variables in terms of the other so that the number of unknowns are atleast reduced. Since step 3 calls for expressing the function with other known quantities, there would be no other way for this to happen than to look for known quantities which are given in the problem. It is stated that the total cost should be at$2400 and we can break down the cost as the sum of the cost of the area of the base and the cost of the sides. The cost of the base C_b, is price times area, thus, C_b=$8*s^2 while the cost of the sides is C_s=4($2*h*s) since there are four sides. Thus, if C is the total cost, $2400=C=C_b+C_s or alternatively$2400=8*s^2+4(\$2*h*s). Thus, we can express s or h in terms of h or s, respectively. We can choose which expresses which but let us choose h such that h=\frac{300-s^2}{s}.
Substituting h to our original expression of the volume, we have

V=s^2*h=s^2(\frac{300-s^2}{s})=300s-s^3

#### Step 4: Finding for the critical values.

To find fof the critical values, we differentiate our function V, equate to zero and then solve for , thus

D_{s}V=\frac{\partial }{\partial s}(300s-s^3)=0
=300-3s^2=0

Thus, solving for s, we have s=\pm 10. But we take only the positve s=10 since there is no negative dimension. Thus, the only critical point of V is at s=10.

#### Step 5: Second derivative and final evaluation.

We know that the first derivative is V=300-3s^2=0. We'll just differentiate this again to get the second derivative which is just

V''(s)=-6s=0
.

Finding for the value at the critical points,

V''(10)=-6(10)=-60

Since the value of the derivative at the critical point is negative, we can safely say that the maximum value of the function would be when s=10 as stipulated in step 5 above. This means that the highest possible volume the box can be made is when the sides at the bottom is 10 feet. Also, we can now solve for the height h which is just given above,

h=\frac{300-s^2}{s}
h=\frac{300-(10)^2}{10}=20

Since we already have the values for our height and sides, we can now solve for the maximum volume which is just given by

V=s^2h=(10)^2*20=2000ft^3.

### Example #1

Find the point on the curve y = 2x^2 that has a minimum distance from the point (9 , 0).

### Example #2

From example #1, what is the minimum distance?

### Example #3

A 2400 sq. m. field (rectangle shape) is to be fenced. Then the owner decided to divide the field into two with another fence that is set anti-parallel to one of the sides. Solve for the dimension of the field that gives the minimum amount of fencing material?

Answer: 40 meters by 60 meters.

### Example #4

In the above example, given the dimension, what is the minimum amount of material needed?