Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let VAB be the volume of the largest cone contained in the sphere.
Obviously, for maximum volume, the axis of the cone should be along the height of the sphere.
Let, ∠AOC = θ
∴ AC, the radius of the base of the cone = R sin θ, where R
is the radius of the sphere.
Height of the cone VC = VO + OC = R + R cos θ
Volume of a cone; V = `1/3 pi (AC)^2 xx (VC)`
`=> V = 1/3 piR^2 sin^2 θ (R + R cos theta)`
`=> V = 1/3 piR^3 sin^2 theta (1 + cos theta)`
On differentiating
`therefore dV/(d theta) = 1/3 piR^3 [sin^2 theta (- sin theta) + (1 cos theta) * 2 sin theta cos theta]`
`= 1/3 piR^3 [- sin^3 theta + 2 sin theta cos theta + 2 sin theta (1 - sin^2 theta)]`
`= 1/3 piR^3 [- sin^3 theta + 2 sin theta cos theta + 2 sin theta - 2sin^2 theta]`
`= 1/3 pi R^3 [- 3 sin^3 theta + 2 sin theta + 2 sin theta cos theta]`
For minimum and maximum, `(dV)/(d theta) = 0`
`=> 1/3 pi"R"^3 (- 3 sin^3 theta + 2 sin cos theta + 2 sin theta)` = 0
= - 3 sin3 θ + 2 sin θ cos θ + 2 sin θ = 0
= sin θ (- 3 sin2 θ + 2 cos θ + 2) = 0
= - 3 sin2 θ + 2 cos θ +2 = 0 ...[∵ sin θ ≠ 0]
= -3 (1 - cos2 θ) + 2 cos θ + 2 = 0
= - 3 + cos2 θ + 2 cos θ + 2 = 0
⇒ 3 cos2 θ + 2 cos θ - 1 = 0
⇒ (3 cos θ - 1)(cos θ + 1) = 0
⇒ cos θ = `1/3` cos θ = - 1
But cos θ ≠ 1 because cos θ = - 1 ⇒ θ = π which is not possible.
`therefore cos theta = 1/3`
When `cos theta = 1/3`, then `sin theta = sqrt(1 - cos^2 theta) = sqrt(1 - 1/9)`
`= sqrt(8/9)`
`= (2 sqrt2)/3`
Now `(dV)/(d theta) = 1/3 piR^3 sin theta [- 3 sin^2 theta + 2 + 2 cos theta]`
`= 1/3 piR^3 sin theta (3 cos theta - 1)(cos theta + 1)`
The sign of `cos theta = 1/3, (dV)/(d theta)` changes from +ve to -ve.
∴ V is highest at `theta = cos^-1 (1/3)`.
On decreasing θ, cos θ increases.
Now cos θ = `1/3` then V is maximum.
The height of the cone for this value of cos θ is
VC = R + R cos θ
`= R + R xx 1/3 = (4R)/3`
Radius of cone = AC = R sin θ = `R * (2sqrt2)/3 = (2 sqrt2)/3 R`
∴ Maximum volume of the cone is V
`= 1/3 pi (AC)^2 (VC)`
`= 1/3 pi ((2 sqrt(2R))/3)^2 ((4R')/3)`
`= 1/3 pi xx 8/9 R^2 xx (4R)/3`
`= 8/27 (4/3 piR^3)`
`= 8/27 xx` Volume of sphere
APPEARS IN
संबंधित प्रश्न
Find the maximum and minimum value, if any, of the function given by f(x) = |x + 2| − 1.
Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:
`h(x) = sinx + cosx, 0 < x < pi/2`
Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:
f(x) = sinx − cos x, 0 < x < 2π
Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:
`g(x) = x/2 + 2/x, x > 0`
Find the local maxima and local minima, if any, of the following function. Find also the local maximum and the local minimum values, as the case may be:
`g(x) = 1/(x^2 + 2)`
Find the absolute maximum value and the absolute minimum value of the following function in the given interval:
f (x) = (x −1)2 + 3, x ∈[−3, 1]
Find two numbers whose sum is 24 and whose product is as large as possible.
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
Show that the right circular cone of least curved surface and given volume has an altitude equal to `sqrt2` time the radius of the base.
The point on the curve x2 = 2y which is nearest to the point (0, 5) is ______.
Find the absolute maximum and minimum values of the function f given by f (x) = cos2 x + sin x, x ∈ [0, π].
Show that a cylinder of a given volume, which is open at the top, has minimum total surface area when its height is equal to the radius of its base.
Show that the height of a cylinder, which is open at the top, having a given surface area and greatest volume, is equal to the radius of its base.
Solve the following : Show that a closed right circular cylinder of given surface area has maximum volume if its height equals the diameter of its base.
Solve the following : Show that the height of a right circular cylinder of greatest volume that can be inscribed in a right circular cone is one-third of that of the cone.
Determine the maximum and minimum value of the following function.
f(x) = x log x
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?
A metal wire of 36 cm long is bent to form a rectangle. By completing the following activity, find it’s dimensions when it’s area is maximum.
Solution: Let the dimensions of the rectangle be x cm and y cm.
∴ 2x + 2y = 36
Let f(x) be the area of rectangle in terms of x, then
f(x) = `square`
∴ f'(x) = `square`
∴ f''(x) = `square`
For extreme value, f'(x) = 0, we get
x = `square`
∴ f''`(square)` = – 2 < 0
∴ Area is maximum when x = `square`, y = `square`
∴ Dimensions of rectangle are `square`
The function y = 1 + sin x is maximum, when x = ______
Find all the points of local maxima and local minima of the function f(x) = `- 3/4 x^4 - 8x^3 - 45/2 x^2 + 105`
If x is real, the minimum value of x2 – 8x + 17 is ______.
The function f(x) = 2x3 – 3x2 – 12x + 4, has ______.
The maximum value of sin x . cos x is ______.
The maximum value of `(1/x)^x` is ______.
The maximum value of `["x"("x" − 1) + 1]^(1/3)`, 0 ≤ x ≤ 1 is:
Find both the maximum and minimum values respectively of 3x4 - 8x3 + 12x2 - 48x + 1 on the interval [1, 4].
Find the area of the largest isosceles triangle having a perimeter of 18 meters.
The point on the curve `x^2 = 2y` which is nearest to the point (0, 5) is
Read the following passage and answer the questions given below.
|
In an elliptical sport field the authority wants to design a rectangular soccer field with the maximum possible area. The sport field is given by the graph of `x^2/a^2 + y^2/b^2` = 1. |
- If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x.
- Find the critical point of the function.
- Use First derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.
OR
Use Second Derivative Test to find the length 2x and width 2y of the soccer field (in terms of a and b) that maximize its area.
If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to ______.
If the function y = `(ax + b)/((x - 4)(x - 1))` has an extremum at P(2, –1), then the values of a and b are ______.
The set of values of p for which the points of extremum of the function f(x) = x3 – 3px2 + 3(p2 – 1)x + 1 lie in the interval (–2, 4), is ______.
A cone of maximum volume is inscribed in a given sphere. Then the ratio of the height of the cone to the diameter of the sphere is ______.
The sum of all the local minimum values of the twice differentiable function f : R `rightarrow` R defined by
f(x) = `x^3 - 3x^2 - (3f^('')(2))/2 x + f^('')(1)`
If \[\mathrm{A}+\mathrm{B}=\frac{\pi}{2}\] then the maximum value of cosA.cosB is
For a function defined on an interval \[I\], which condition means that \[f\] has a minimum value at \[c\in I\]?
A maximum or minimum value of \[f\] is called a(n) _____.
