Advertisements
Advertisements
प्रश्न
If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sum of their volumes is minimum?
Advertisements
उत्तर
Let x be the edge of the cube and r be the radius of the sphere.
Surface area of cube = 6x2
And surface area of the sphere = 4πr2
∴ 6x2 + 4πr2 = K ......(constant)
⇒ r = `sqrt(("K" - 6x^2)/(4pi)` .....(i)
Volume of the cube = x3 and the volume of sphere = `3/4 pi"r"^3`
∴ Sum of their volumes (V) = Volume of cube + Volume of sphere
⇒ V = `x^3 + 4/3 pi"r"^3`
⇒ V = `x^3 + 4/3 pi xx (("K" - 6x^2)/(4pi))^(3/2)`
Differentiating both sides w.r.t. x, we get
`"dV"/"dx" = 3x^2 + (4pi)/3 xx 3/2("K" - 6x^2)^(1/2) (- 12x) xx 1/((4pi)^(3/2)`
= `3x^2 + (2pi)/((4pi)^(3/2)) xx (-12x) ("K" - 6x^2)^(1/2)`
= `3x^2 + 1/(4pi^(1/2)) xx (-12x) ("K" - 6x^2)^(1/2)`
∴ `"dV"/"dx" = 3x^2 - (3x)/sqrt(pi) ("K" - 6x^2)^(1/2)` ....(ii)
For local maxima and local minima, `"dV"/"dx"` = 0
∴ `3x^2 - (3x)/sqrt(pi) ("K" - 6x^2)^(1/2)` = 0
⇒ `3x[x - ("k" - 6x^2)^(1/2)/sqrt(pi)]` = 0
x ≠ 0
∴ `x - ("K" - 6x^2)^(1/2)/sqrt(pi)` = 0
⇒ x = `("K" - 6x^2)^(1/2)/sqrt(pi)`
Squaring both sides, we get
x2 = `("K" - 6x^2)/pi`
⇒ `pix^2 = "K" - 6x^2`
⇒ `pix^2 + 6x^2` = K
⇒ `x^2(pi + 6)` = K
⇒ x2 = `"K"/(pi + 6)`
∴ x = `sqrt("K"/(pi + 6)`
Now putting the value of K in equation (i), we get
`6x^2 + 4pir^2 = x^2(pi + 6)`
⇒ `6x^2 + 4pi"r"^2 = pix^2 + 6x^2`
⇒ `4pi"r"^2 = pi"r"^2`
⇒ 4r2 = x2
∴ 2r = x
∴ x:2r = 1:1
Now differentiating equation (ii) w.r.t x, we have
`("d"^2"V")/("dx"^2) = 6x - 3/sqrt(pi) "d"/"dx" [x("K" - 6x^2)^(1/2)]`
= `6x - 3/sqrt(pi)[x * 1/(2sqrt("K" - 6x^2)) xx (-12x) + ("K" - 6x^2)^(1/2) * 1]`
= `6x - 3/sqrt(pi) [(-6x^2)/sqrt("K" - 6x^2) + sqrt("K" - 6x^2)]`
= `6x - 3/sqrt(pi) [(-6x^2 + "K" - 6x^2)/sqrt("K" - 6x^2)]`
= `6x + 3/sqrt(pi) [(12x^2 - "K")/sqrt("K" - 6x^2)]`
Put x = `sqrt("K"/(pi + 6)`
= `6sqrt("K"/(pi + 6)) + 3/sqrt(pi)[((12"K")/(pi + 6) - "K")/sqrt("K" - (6"K")/(pi + 6))]`
= `6sqrt("K"/(pi + 6)) + 3/sqrt(pi) [(12"K" - pi"K" - 6"K")/sqrt((pi"K" + 6"K" - 6"K")/(pi + 6))]`
= `6sqrt("K"/(pi + 6)) + 3/sqrt(pi) [(6"K" - pi"K")/sqrt((pi"K")/(pi + 6))]`
= `6sqrt("K"/(pi + 6)) + 3/(pisqrt("K"))[(6"K" - pi"K") sqrt(pi + 6)] > 0`
So it is minima.
Hence, the required ratio is 1 : 1 when the combined volume is minimum.
APPEARS IN
संबंधित प्रश्न
Examine the maxima and minima of the function f(x) = 2x3 - 21x2 + 36x - 20 . Also, find the maximum and minimum values of f(x).
If the sum of lengths of hypotenuse and a side of a right angled triangle is given, show that area of triangle is maximum, when the angle between them is π/3.
Find the maximum and minimum value, if any, of the following function given by g(x) = x3 + 1.
Find the absolute maximum value and the absolute minimum value of the following function in the given interval:
`f(x) =x^3, x in [-2,2]`
Find the absolute maximum value and the absolute minimum value of the following function in the given interval:
f (x) = sin x + cos x , x ∈ [0, π]
At what points in the interval [0, 2π], does the function sin 2x attain its maximum value?
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Find two positive numbers x and y such that x + y = 60 and xy3 is maximum.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening
A rod of 108 meters long is bent to form a rectangle. Find its dimensions if the area is maximum. Let x be the length and y be the breadth of the rectangle.
Divide the number 30 into two parts such that their product is maximum.
Solve the following : Show that of all rectangles inscribed in a given circle, the square has the maximum area.
Divide the number 20 into two parts such that their product is maximum.
Twenty meters of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is ______
A telephone company in a town has 500 subscribers on its list and collects fixed charges of Rs 300/- per subscriber per year. The company proposes to increase the annual subscription and it is believed that for every increase of Re 1/- one subscriber will discontinue the service. Find what increase will bring maximum profit?
AB is a diameter of a circle and C is any point on the circle. Show that the area of ∆ABC is maximum, when it is isosceles.
The maximum value of sin x . cos x is ______.
The maximum value of `(1/x)^x` is ______.
The curves y = 4x2 + 2x – 8 and y = x3 – x + 13 touch each other at the point ______.
If y = x3 + x2 + x + 1, then y ____________.
Find both the maximum and minimum values respectively of 3x4 - 8x3 + 12x2 - 48x + 1 on the interval [1, 4].
A ball is thrown upward at a speed of 28 meter per second. What is the speed of ball one second before reaching maximum height? (Given that g= 10 meter per second2)
The point on the curve `x^2 = 2y` which is nearest to the point (0, 5) is
The minimum value of α for which the equation `4/sinx + 1/(1 - sinx)` = α has at least one solution in `(0, π/2)` is ______.
If y = alog|x| + bx2 + x has its extremum values at x = –1 and x = 2, then ______.
The function g(x) = `(f(x))/x`, x ≠ 0 has an extreme value when ______.
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 lateral edge of a regular rectangular pyramid is 'a' cm long. The lateral edge makes an angle a. with the plane of the base. The value of a for which the volume of the pyramid is greatest, is ______.
The minimum value of 2sinx + 2cosx is ______.
A metal wire of 36 cm long is bent to form a rectangle. Find its dimensions when its area is maximum.
Complete the following activity to divide 84 into two parts such that the product of one part and square of the other is maximum.
Solution: Let one part be x. Then the other part is 84 - x
Letf (x) = x2 (84 - x) = 84x2 - x3
∴ f'(x) = `square`
and f''(x) = `square`
For extreme values, f'(x) = 0
∴ x = `square "or" square`
f(x) attains maximum at x = `square`
Hence, the two parts of 84 are 56 and 28.
A running track of 440 m is to be laid out enclosing a football field. The football field is in the shape of a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum,then find the length of its sides. Also calculate the area of the football field.
A right circular cylinder is to be made so that the sum of the radius and height is 6 metres. Find the maximum volume of the cylinder.
Sumit has bought a closed cylindrical dustbin. The radius of the dustbin is ‘r' cm and height is 'h’ cm. It has a volume of 20π cm3.
- Express ‘h’ in terms of ‘r’, using the given volume.
- Prove that the total surface area of the dustbin is `2πr^2 + (40π)/r`
- Sumit wants to paint the dustbin. The cost of painting the base and top of the dustbin is ₹ 2 per cm2 and the cost of painting the curved side is ₹ 25 per cm2. Find the total cost in terms of ‘r’, for painting the outer surface of the dustbin including the base and top.
- Calculate the minimum cost for painting the dustbin.
The shortest distance between the line y - x = 1and the curve x = y2 is
