Lesson Calculus optimization problems for shapes in 2D plane

Calculus optimization problems for shapes in 2D plane

Problem 1

If x is the length of the garden, then its width is   feet, and the area is


    A(x) =  = 


They want you find the maximum of the function A(x) using Calculus.


For it, differentiate A(x) over x and equate the derivative to zero:


    A'(x) = -2x + 50 = 0,


which gives you  x =  = 25 feet.


Thus you obtain the


ANSWER.  Under given conditions, the maximum area is achieved for the square of the side length equal to 

         one fourth of the given perimeter, i.e.  25 feet.

Problem 2

Let x be the dimension of the pen (in feet) perpendicular to the wall. 


Then the side parallel to the wall is  (500-2x)  feet long,


and the area of the pen is


    A(x) = x*(500-2x)  = -2x^2 + 500x  square feet.


They want you find the maximum of the function A(x) using Calculus.


For it, differentiate A(x) over x and equate the derivative to zero:


    A'(x) = -4x + 500 = 0,


which gives you  x =  = 125 feet.


Thus you obtain the


ANSWER.  Under given conditions, the maximum area is achieved for the rectangle 

         with the short side of 125 ft perpendicular to the wall and long side of 500 - 2*125 = 250 ft parallel to the wall.

Problem 3

Let x be the coral length in the direction parallel to the river and y be the coral width in the perpendicular direction.



Then your task is to minimize  (x+2y)  under the given condition/restriction  on the area

    xy = 1800                (1)



From (1),  x = ,  so we need to minimize the function  f(y) = .


The derivative  f'(y) = - + 2.


To find the minimum of f(y), equate its derivative to zero


    - + 2 = 0


     = 2

     =  = 900

    y =  = 30.


ANSWER.  The minimum fencing is at y = 30 ft perpendicular to the river and x =  =  = 60 ft along the river.

Problem 4

If x*y = 100,  then  y = ,  and the sum  x+y  is  x + .


This function of "x",  f(x) = x +   achieves the minimum when its derivative is equal to zero


    f'(x) = 1 -  = 0.


Then    = 100;  hence  x =  = 10.


Thus we proved that if  x*y = 100,  then  x+y has minimum at  x = y = 10.

Problem 5

From the context, the dimensions of the rectangle are not given for advance - they are unknowns and they should be found

from the minimum cost condition.


Let x be one dimension and y be the other dimension of the rectangle.


Then the cost of the outside perimeter fence is 1.50*(2x+2y) dollars = 3*(x+y) dollars,

while the cost of the fence down middle is 0.50*x dollars.


    Note, that I don't know now, which dimension will be the length and which be the width.

    When the problem will be solved, the solution will tell me it . . . 


So, I need minimize the function  

        f(x,y) = 3*(x+y) + 0.5x     (1)

under the condition

        x*y = 1050.                 (2)



From (2), express  y =   and substitute it into (1). You will get

        g(x) =  =  +  + .


Differentiate it over x

        g'(x) =  -  + 

and equate the derivative to zero.  You will get


        3.5 = ,    or

        3.5x^2 = 3150

           x^2 =  = 900,

           x   =  = 30.


Thus the dimensions of the rectangle are 30 ft  and   = 35 ft.

The fence down middle has the length of 30 ft;  hence, it is parallel to the shorter side of the rectangle.

Problem 6

Let "x" be the square side length, and "y" be the radius of the circle.

Then

    4x +  = 16 inches                               (1)  (perimeter)

    f(x,y) = x^2 +    is the function to minimize   (2)  (area)


In other words, you should minimize (2) under the constraint (1).


From (1), express x =  =   and substitute it into the function (2).

You will find then the function to minimize in the form

    g(y) =  +  = 

         =  +  = .


This quadratic function of "y" has the minimum at  y = "  " =  = .    


ANSWER.  The values that provide the minimum of the total area are

         y=  inches (the circle radius)  and  

         x =  =  =  =  = 2.4 inches (the side of the square).

Problem 7

They want you to minimize

    f(x,y) = (x-6)*(y-4)  under the condition  x*y = 96.  (1)

where x and y are the dimensions of the poster.


In other words, from (1), you need to minimize

    f(x,y) = xy - 4x - 6y + 24 = 96 - 4x - 6y + 24 = -4x - 6y + 120.


From (1), you have  y = , so the function f(x,y) takes the form

    g(x) = -4x -  + 120 = -4x -  + 120.    (2)


So, you differentiate (2), and you get

    g'(x) = -4 + .


Equate it to zero

    g'(x) = 0 = -4 + .


So, to find x, you have this equation

     = 576,

     =  = 144

     x =  = 12 centimeters.


Then y =  = 8 cm.


Thus the dimension of the poster are  12 inches width and 8 inches height.

Problem 8

If  x+y = 9,  then  y = 9-x,  and the function  f(x,y) =   is


    f(x,y) =  = .


So, we need to find the maximum of the function of x  g(x) = .


To find the maximum of the function g(x), take its derivative and equate it to zero.


The derivative is

    g'(x) =  - 2x*(9-x) =  -  = .


Equating it to zero, you get

     = 0,  which is equivalent to   = 0.


Left side is factorable

    (x-9)*(x-3) = 0.


and two solution of the quadratic equation are x= 9  and  x= 3.


The value x= 9 provides the local minimum of the function f(x), while x= 3 provides the local maximum.


At x= 3,  y= 9-3 = 6,  and the function g(x,y) =  = 108.



    


        Plot  g(x) = x*(9-x)^2