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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.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.

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

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.

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.

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