Advertisements
Advertisements
प्रश्न
A box with a square base is to have an open top. The surface area of the box is 192 sq cm. What should be its dimensions in order that the volume is largest?
Advertisements
उत्तर
Let x cm be the side of square base and h cm be its height.
Then x2 + 4xh = 192
∴ h = `(192 - x^2)/(4x)` ...(1)
Let V = `x^2"h"`
= `x^2((192 - x^2)/(4x))` ...[By (1)]
∴ V = `(1)/(4)(192x - x^3)`
∴ `"dV"/("d"x) = (1)/(4) "d"/("d"x)(192x - x^3)`
= `(1)/(4)(192 xx 1 - 3x^2)`
= `(3)/(4)(64 - x^2)`
and
`("d"^2"V")/("d"x^2) - (3"d")/(4"d"x)(64 - x^2)`
= `(3)/(4)(0 - 2x)`
= `-(3)/(2)x`
For maximum V, `"dV"/("d"x)` = 0
∴ `(3)/(4)(64 - x^2)` = 0
∴ x2 = 64
∴ x = 8 ...[∵ x > 0]
and
`(("d"^2V)/("d"x^2))_("at" x = 8)`
= `-(3)/(2) xx 8`
= – 12 < 0
∴ by the second derivative test, V is maximum at x = 8.
If x = 8,
h = `(192 - 64)/(4(8)`
= `(128)/(32)`
= 4
Hence, the volume of the box is largest, when the side of square base is 8 cm and its height is 4 cm.
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).
Find the maximum and minimum value, if any, of the following function given by f(x) = −(x − 1)2 + 10
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 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 maximum profit that a company can make, if the profit function is given by p(x) = 41 − 72x − 18x2.
Find the maximum and minimum values of x + sin 2x on [0, 2π].
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.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is `tan^(-1) sqrt(2)`
The maximum value of `[x(x −1) +1]^(1/3)` , 0 ≤ x ≤ 1 is ______.
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.`
Prove that the semi-vertical angle of the right circular cone of given volume and least curved surface is \[\cot^{- 1} \left( \sqrt{2} \right)\] .
A given quantity of metal is to be cast into a half cylinder with a rectangular base and semicircular ends. Show that in order that the total surface area may be minimum the ratio of the length of the cylinder to the diameter of its semi-circular ends is \[\pi : (\pi + 2)\].
Find the maximum and minimum of the following functions : f(x) = x3 – 9x2 + 24x
A wire of length 36 metres is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum.
Choose the correct option from the given alternatives :
If f(x) = `(x^2 - 1)/(x^2 + 1)`, for every real x, then the minimum value of f is ______.
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:
A wire of length l is cut into two parts. One part is bent into a circle and the other into a square. Show that the sum of the areas of the circle and the square is the least, if the radius of the circle is half of the side of the square.
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) = x log x
A wire of length 120 cm is bent in the form of a rectangle. Find its dimensions if the area of the rectangle is maximum
A rod of 108 m long is bent to form a rectangle. Find it’s dimensions when it’s area is maximum.
Max value of z equals 3x + 2y subject to x + y ≤ 3, x ≤ 2, -2x + y ≤ 1, x ≥ 0, y ≥ 0 is ______
If R is the circum radius of Δ ABC, then A(Δ ABC) = ______.
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 ______
The sum of two non-zero numbers is 6. The minimum value of the sum of their reciprocals is ______.
Let f have second derivative at c such that f′(c) = 0 and f"(c) > 0, then c is a point of ______.
The sum of the surface areas of a rectangular parallelopiped with sides x, 2x and `x/3` and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.
The function f(x) = 2x3 – 3x2 – 12x + 4, has ______.
Maximum slope of the curve y = –x3 + 3x2 + 9x – 27 is ______.
The maximum value of `["x"("x" − 1) + 1]^(1/3)`, 0 ≤ x ≤ 1 is:
Find all the points of local maxima and local minima of the function f(x) = (x - 1)3 (x + 1)2
If y `= "ax - b"/(("x" - 1)("x" - 4))` has a turning point P(2, -1), then find the value of a and b respectively.
The area of a right-angled triangle of the given hypotenuse is maximum when the triangle is ____________.
Find the area of the largest isosceles triangle having a perimeter of 18 meters.
The coordinates of the point on the parabola y2 = 8x which is at minimum distance from the circle x2 + (y + 6)2 = 1 are ____________.
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 function `f(x) = x^3 - 6x^2 + 9x + 25` has
Check whether the function f : R `rightarrow` R defined by f(x) = x3 + x, has any critical point/s or not ? If yes, then find the point/s.
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 point on the curve y2 = 4x, which is nearest to the point (2, 1).
