Advertisements
Advertisements
Question
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
Solution
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
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.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is `(4r)/3`. Also find maximum volume in terms of volume of the sphere
Find the maximum and minimum value, if any, of the following function given by h(x) = sin(2x) + 5.
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) = x2
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`
Prove that the following function do not have maxima or minima:
f(x) = ex
Find the maximum value of 2x3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube.
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.
The volume of a closed rectangular metal box with a square base is 4096 cm3. The cost of polishing the outer surface of the box is Rs. 4 per cm2. Find the dimensions of the box for the minimum cost of polishing it.
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.
A rectangle is inscribed in a semicircle of radius r with one of its sides on the diameter of the semicircle. Find the dimensions of the rectangle to get the maximum area. Also, find the maximum area.
Find the maximum and minimum of the following functions : f(x) = x3 – 9x2 + 24x
Find the maximum and minimum of the following functions : f(x) = x log x
Find the largest size of a rectangle that can be inscribed in a semicircle of radius 1 unit, so that two vertices lie on the diameter.
Show that the height of a closed right circular cylinder of given volume and least surface area is equal to its diameter.
Solve the following:
A rectangular sheet of paper of fixed perimeter with the sides having their lengths in the ratio 8 : 15 converted into an open rectangular box by folding after removing the squares of equal area from all corners. If the total area of the removed squares is 100, the resulting box has maximum volume. Find the lengths of the rectangular sheet of paper.
Solve the following : Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is `(2"R")/sqrt(3)`. Also, find the maximum volume.
Determine the maximum and minimum value of the following function.
f(x) = 2x3 – 21x2 + 36x – 20
State whether the following statement is True or False:
An absolute maximum must occur at a critical point or at an end point.
Find the local maximum and local minimum value of f(x) = x3 − 3x2 − 24x + 5
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`
If z = ax + by; a, b > 0 subject to x ≤ 2, y ≤ 2, x + y ≥ 3, x ≥ 0, y ≥ 0 has minimum value at (2, 1) only, then ______.
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`
The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is ______.
Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is ______.
Let f(x) = 1 + 2x2 + 22x4 + …… + 210x20. Then f (x) has ____________.
The function `f(x) = x^3 - 6x^2 + 9x + 25` has
Divide 20 into two ports, so that their product is maximum.
A function f(x) is maximum at x = a when f'(a) > 0.
If S1 and S2 are respectively the sets of local minimum and local maximum points of the function. f(x) = 9x4 + 12x3 – 36x2 + 25, x ∈ R, then ______.
If the point (1, 3) serves as the point of inflection of the curve y = ax3 + bx2 then the value of 'a ' and 'b' are ______.
The maximum distance from origin of a point on the curve x = `a sin t - b sin((at)/b)`, y = `a cos t - b cos((at)/b)`, both a, b > 0 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"`
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) `= x sqrt(1 - x), 0 < x < 1`
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.
20 is divided into two parts so that the product of the cube of one part and the square of the other part is maximum, then these two parts are
