PROBLEMS INVOLVING MAXIMUM AND MINIMUM VALUIES IN DERIVATIVES

Optimization is a process of finding an extreme value (either maximum or minimum) under certain conditions.

A procedure for solving for an extremum or optimization problems.

Step 1 :

Draw an appropriate figure and label the quantities relevant to the problem.

Step 2 :

Find a expression for the quantity to be maximized or minimized.

Step 3 :

Using the given conditions of the problem, the quantity to be extremized .

Step 4 :

Determine the interval of possible values for this variable from the conditions given in the problem.

Step 5 :

Using the techniques of extremum (absolute extremum, first derivative test or second derivative test) obtain the maximum or minimum.

Problem 1 :

Find the dimensions of the largest rectangle that can be inscribed in a semi circle of radius r cm.

Solution :

Let θ be the angle made by OP with the positive direction of x axis.

Area of rectangle A(θ)  =  (2 r cos θ) (r sinθ)

=  r2 sin 2θ

A'(θ)  =  2r2 cos 2θ

A'(θ)  =  0

2r2 cos 2θ  =  0

2θ  =  cos-1(0)

2θ  =  π/2

θ  =  π/4

A''(θ)  =  -4r2 sin 2θ

A''(π/4)  =  -4r2 sin 2(π/4)

A''(π/4)  =  -4 r2 < 0 (Maximum)

Length of the rectangle  =  2r cos θ  =  2rcos(π/4)

=  2 r (1/√2)

=  2 r (1/√2)

=  √2r

Width of the rectangle  =  r sin θ

=  r sin (π/4)

=  r (1/√2)

=  r/√2

Problem 2 :

A manufacturer wants to design an open box having a square base and a surface area of 108 sq.cm. Determine the dimensions of the box for the maximum volume.

Solution :

Let x be the side length of square base box.

Volume of the box  =  length x width x height

Let h be the height 

Surface area of box  =  x2 + 4xh

x2 + 4xh  =  108

4xh  =  108-x2

h  =  (108-x2)/4x  ---(1)

Volume of the box  =  Base area x height

V(x)  =  x2 (h)  ---(2)

V(x)  =  x2(108-x2)/4x

V(x)  =  (108x-x3)/4

V'(x)  =  (108 - 3x2)/4

V'(x)  =  0

(108 - 3x2)/4  =  0

3x2  =  108

x2  =  36

x  =  6

V''(x)  =  -6x/4

V''(x)  =  -3x/2

V''(6)  =  -3(6)/2

V''(6)  =  -9 < 0 (Maximum)

Applying x  =  6 in (1), we get

h  =  (108-x2)/4x

h  =  (108-36)/24

h  = 3 cm

So, side length of the base is 4 cm and height is 3 cm.

Problem 3 :

The volume of a cylinder is given by the formula

V  =  π r2h

Find the greatest and least values of V if r + h = 6.

Solution :

Given :

V  =  π r2h and r + h  =  6

h  =  6 - r

V(r)  =  π r2(6-r)

V(r)  =  π(6r2-r3)

V'(r)  =  π(12r-3r2)

V'(r)  =  0

3rπ(4-r)  =  0

r  =  0 and r  =  4

V''(r)  =  π(12-6r)

V''(0)  =  π(12-6(0))

V''(0)  =  12π > 0

Minimum

V''(r)  =  π(12-6r)

V''(4)  =  π(12-6(4))

V''(0)  =  -12π < 0

Maximum

r+h  =  6

Case 1 : When r  =  0, h  =  6

Case 2 : When r  =  4, h  =  2

V  =  π r2h

By applying case 1,  V  =  0

By applying case 2,  V  =  π (4)2(2)  ==>  32π

So, the greatest value of V is 32π and least value of V is 0.

Problem 4 :

A hollow cone with base radius a cm and height b cm is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is 4/times volume of the cone.

Solution :

Volume of cone  =  (1/3)π r2h----(1)

Here r  =  a and h  =  b

=  (1/3)π r2h

=  (1/3)π a2b

OAB and CDB are similar.

OA/DC  =  OB/DB

b/h  =  a/(a-r)

h  =  (b/a)(a-r)

V(r)  =  (1/3)π r2[(b/a)(a-r)]

V(r)  =  (b/3a)π r2(a-r)

V(r)  =  (bπ/3a) (ar2-r3)

V'(r)  =  (bπ/3a) (2ar-3r2)

V'(r)  =  (bπ/3a) (2ar-3r2)  =  0

r(2a-3r)  =  0

r  =  0, r  =  2a/3

V''(r)  =  (bπ/3a) (2a-6r)

V''(0)  =  (bπ/3a) (2a) > 0 Minimum

V''(r)  =  (bπ/3a) (2a-6(2a/3))

V''(2a/3)  =  (bπ/3a) (2a-6(2a/3)) < 0 Maximum

Volume of cylinder  =  πr2h

r  =  2a/3 and h  =  (b/a)(a-(2a/3))

h  =  (b/a) (a/3)

h  =  b/3

=  π(2a/3)2(b/3)

Volume of cylinder  =  (4/9)(1/3)πa2b

Volume of cylinder  =  (4/9)Volume of cone

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