English

A rod of 108 m long is bent to form a rectangle. Find it’s dimensions when it’s area is maximum. - Mathematics and Statistics

Advertisements
Advertisements

Question

A rod of 108 m long is bent to form a rectangle. Find it’s dimensions when it’s area is maximum.

Sum

Solution

Let the length and breadth of a rectangle be l and b respectively.

∴ Perimeter of rectangle = 2(l + b) = 108 m

∴ l + b = 54

∴ b = 54 – l     ...(i)

Area of rectangle = l × b

= l(54 – l)    ...[From (i)]

Let f(l) = l(54 – l)

= 54l – l2

∴ f'(l) = 54 – 2l

∴ f"(l) = – 2

Consider, f'(l) = 0

∴ 54 – 2l = 0

∴ 54 = 2l

∴ l = 27

For l = 27,

f"(27) = – 2 < 0

∴ f(l) i.e., area is maximum at l = 27

and b = 54 – 27    ...[From (i)]

= 27

∴ The dimensions of rectangle are 27 m × 27 m.

shaalaa.com
  Is there an error in this question or solution?
Chapter 1.4: Applications of Derivatives - Q.5

RELATED QUESTIONS

Show that the height of the cylinder of maximum volume, that can be inscribed in a sphere of radius R is `(2R)/sqrt3.`  Also, find the maximum volume.


Find the maximum and minimum value, if any, of the following function given by g(x) = − |x + 1| + 3.


Find the maximum and minimum value, if any, of the following function given by h(x) = x + 1, x ∈ (−1, 1)


Prove that the following function do not have maxima or minima:

f(x) = ex


Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.


Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?


Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is `8/27` of the volume of the sphere.


Find the points at which the function f given by f (x) = (x – 2)4 (x + 1)3 has

  1. local maxima
  2. local minima
  3. point of inflexion

Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to \[ \frac{2}{3} \] of the diameter of the sphere.


 Find the point on the straight line 2x+3y = 6,  which is closest to the origin. 


Find the maximum and minimum of the following functions : y = 5x3 + 2x2 – 3x.


Determine the maximum and minimum value of the following function.

f(x) = 2x3 – 21x2 + 36x – 20


Divide the number 20 into two parts such that their product is maximum.


A rectangular sheet of paper has it area 24 sq. Meters. The margin at the top and the bottom are 75 cm each and the sides 50 cm each. What are the dimensions of the paper if the area of the printed space is maximum?


If R is the circum radius of Δ ABC, then A(Δ ABC) = ______.


The function y = 1 + sin x is maximum, when x = ______ 


If the sum of the lengths of the hypotenuse and a side of a right-angled triangle is given, show that the area of the triangle is maximum when the angle between them is `pi/3`


If x is real, the minimum value of x2 – 8x + 17 is ______.


Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is ______.


Find the local minimum value of the function f(x) `= "sin"^4" x + cos"^4 "x", 0 < "x" < pi/2`


Find the points of local maxima and local minima respectively for the function f(x) = sin 2x - x, where `-pi/2 le "x" le pi/2`


If y = x3 + x2 + x + 1, then y ____________.


Find the height of the cylinder of maximum volume that can be inscribed in a sphere of radius a.


The point on the curve `x^2 = 2y` which is nearest to the point (0, 5) is


A function f(x) is maximum at x = a when f'(a) > 0.


Let P(h, k) be a point on the curve y = x2 + 7x + 2, nearest to the line, y = 3x – 3. Then the equation of the normal to the curve at P is ______.


The rectangle has area of 50 cm2. Complete the following activity to find its dimensions for least perimeter.

Solution: Let x cm and y cm be the length and breadth of a rectangle.

Then its area is xy = 50

∴ `y =50/x`

Perimeter of rectangle `=2(x+y)=2(x+50/x)`

Let f(x) `=2(x+50/x)`

Then f'(x) = `square` and f''(x) = `square`

Now,f'(x) = 0, if x = `square`

But x is not negative.

∴ `x = root(5)(2)   "and" f^('')(root(5)(2))=square>0`

∴ by the second derivative test f is minimum at x = `root(5)(2)`

When x = `root(5)(2),y=50/root(5)(2)=root(5)(2)`

∴ `x=root(5)(2)  "cm" , y = root(5)(2)  "cm"`

Hence, rectangle is a square of side `root(5)(2)  "cm"`


Determine the minimum value of the function.

f(x) = 2x3 – 21x2 + 36x – 20


Mrs. Roy designs a window in her son’s study room so that the room gets maximum sunlight. She designs the window in the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 m, find the dimensions of the window that will admit maximum sunlight into the room.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×