# Optimization Problems

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

In this section of ask-math, we are going to see the optimization problems. In optimization problems we are finding the best solutions for all feasible solutions. Here we look for the maximum or minimum value that a function can take. Ask-math already explains you about the absolute extrema where we found the largest and the smallest value of the given function.
Example 1: Find the two numbers whose sum is 24 and whose product is as large as possible.
solution : Let the number be x and y,
x +y = 24 (Given)
y = 24 -x
Let the product be P
P = xy
P = x(24 -x)
P = 24x - $x^{2}$

$\frac{\text{d}P}{\text{d}x}$= 24 - 2x

$\dfrac{d^2 P}{dx^2}$ = -2

For maximum and minimum, we have,

$\frac{\text{d}P}{\text{d}x}$ = 0
24 - 2x = 0
x = 12

Also, $\dfrac{d^2 P}{dx^2}$(at x = 12) = -2 < 0

So x= 12 is a point of maximum.
So when x = 12, y = 12
Hence the required numbers are both equal to 12.

Example 2 : In the right circular cylindrical can which enclose a given volume of 100 cu.cm, which ahs the minimum surface area?
Solution : Let 'r' be the radius and 'h' be the height of the cylindrical can of
Volume = 100 cu.cm
V = $\pi r^{2}$ h
$\pi r^{2}$ h = 100

h = $\frac{100}{\pi r^{2}}$
Let S be the surface area of the can,
S = 2$\pi rh + 2\pi r^{2}$
S = $\frac{200}{r} + 2\pi r^{2}$

$\frac{\text{d}S}{\text{d}r} = \frac{-200}{r^{2}}+ 4 \pi r$

$\dfrac{d^2 S}{dr^2} = \frac{400}{r^{3}} + 4\pi$

For minimum values of S, we have,
$\frac{\text{d}S}{\text{d}r}$ = 0

$\frac{-200}{r^{2}}+ 4 \pi r$ = 0

$4\pi r^{3}$ = 200

r = $(\frac{50}{\pi})^{1/3}$

As $\dfrac{d^2 S}{dr^2}$ > 0

Hence, S is minimum when r = $(\frac{50}{\pi})^{1/3}$

Example 3 : A rectangular page is to contain 20 square inches of print. The margins at the top and bottom of the page are to be 1.5 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the page be so that the least amount of paper is used?
Solution : Let A be the area to be minimized
Let x be the width of the page excluding margin and y be the length excluding margin
Area = ( x+ 3)(y +2) ----(area including margins)
Printed area ,
Area = xy
xy = 20
y = 20/x
Area = ( x+ 3)(20/x +2)
A = 20 + 2x + 60/x + 6
A = 26 + 2x + 60/x
For the minimum,

$\frac{\text{d}A}{\text{d}x} = 0$

2 - $\frac{60}{x^{2}}$ = 0

- $\frac{60}{x^{2}}$ = -2

$\frac{60}{x^{2}}$ = 2

$x^{2}$ = 30
x= $\pm \sqrt{30}$
x= 5.477
x = 5.5
y = 20/x
y = 3.6
So area is minimum when x = 5.5.
dimensions of the page should be x + 3 = 5.5 + 3 = 8.3inches by y + 2= 3.6 + 2 = 5.6inches.

## Guidelines for solving  Optimization Problems

1) Find all the given quantities and the quantities that is to be determined.
2) If possible , draw the diagram.
3) Write the formula according to the given question.
4) Write the formula using the single variable using the given information.
5) Determine the desired maximum or minimum value by the calculus techniques which means find the 1st and 2nd derivative.