MAXIMUM AND MINIMUM VALUE WORD PROBLEMS USING 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 two positive numbers whose sum is 12 and their product is maximum.

Solution

Problem 2 :

Find two positive numbers whose product is 20 and their sum is minimum.

Solution

Problem 3 :

Find the smallest possible value of

x2 + y2

given that x + y =10.

Solution

Problem 4 :

A garden is to be laid out in a rectangular area and protected by wire fence. What is the largest possible area of the fenced garden with 40 meters of wire.

Solution

Problem 5 :

A square sheet of cardboard with each side a centimeters is to be used to make an open top box by cutting a small square of cardboard from each of the corners and bending up the sides. What is the side length of small squares if the box is to have large a volume as possible ?

Solution

Answer Key

1)  The first number is 6 and the another number is 6.

Product of two numbers = 36.

2)  the sum of numbers  =  2√5 + 2√5  ==>  4√5.

3)   the value of x+ y2 is 50.

4)  area of the rectangle is 100 m2.

5)   the maximum volume is 2a3/27

Problem 1 :

A rectangular page is to contain 24 cm2 of print. The margins at the top and bottom of the page are 1.5 cm and the margins at other sides of the page is 1 cm. What should be the dimensions of the page so that the area of the paper used is minimum.

Solution

Problem 2 :

A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq.mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?

Solution

Problem 3 :

Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10 cm.

Solution

Problem 4 :

Prove that among all the rectangles of the given perimeter, the square has the maximum area.

Solution

Problem 5 :

The radius of a sphere is decreasing at a rate of 2 centimeters per second. At the instant when the radius of the sphere is 3 centimeters, what is the rate of change in square centimeters per second, of the surface area of the sphere ?

(The surface area S of a sphere with radius r is 4πr2)

Solution

Problem 6 :

The radius of a circle is increasing at a constant rate of 0.2 meters per second. What is the rate of increase in the area of the circle at the instance when the circumference is 20π meters ?

Solution

Problem 7 :

A box with a square base with no top has a surface area of 108 square feet. Find the dimension that will maximize the volume 

Solution

Answer Key

1)  Length and width of the rectangle is 6 cm and 9 cm.

2)  1200 m

3)  the dimensions are 10√2 cm, 10√2 cm.

4)  It is square

5)  surface area is decreasing at the rate of 48π cm2

6)  the required increasing rate is 4 π.

7)  So, the required dimension of the cube is 6 feet.

Problem 1 :

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

Solution

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

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

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 4/9 times volume of the cone.

Solution

Answer Key

1)  Length = √2r, width = r/√2

2)  side length of the base is 4 cm and height is 3 cm.

3) the greatest value of V is 32π and least value of V is 0.

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

5)  the required volume is 820.84 cm3

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Digital SAT Math Problems and Solutions (Part - 154)

    Apr 30, 25 11:12 PM

    digitalsatmath187.png
    Digital SAT Math Problems and Solutions (Part - 154)

    Read More

  2. Digital SAT Math Problems and Solutions (Part - 153)

    Apr 29, 25 12:18 PM

    digitalsatmath185.png
    Digital SAT Math Problems and Solutions (Part - 153)

    Read More

  3. Digital SAT Math Problems and Solutions (Part - 152)

    Apr 28, 25 11:54 AM

    Digital SAT Math Problems and Solutions (Part - 152)

    Read More