Home
Class 14
MATHS
A cylindrical jar without a lid has to b...

A cylindrical jar without a lid has to be constructed using a given surface area of a metal sheet It the capacity of the jar is to maximum, then the diameter of the jar must be k times the height of the jar The value of k is:

A

1

B

2

C

3

D

4

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of maximizing the capacity of a cylindrical jar without a lid, we will follow these steps: ### Step 1: Understand the Surface Area and Volume Formulas For a cylindrical jar without a lid, the surface area \( S \) and volume \( V \) can be expressed as: - Surface Area: \( S = 2\pi rh + \pi r^2 \) (where \( r \) is the radius and \( h \) is the height) - Volume: \( V = \pi r^2 h \) ### Step 2: Relate Diameter and Height According to the problem, the diameter \( D \) of the jar is \( k \) times the height \( h \). Since the diameter is twice the radius, we have: \[ D = 2r = k \cdot h \] From this, we can express the radius in terms of height: \[ r = \frac{kh}{2} \] ### Step 3: Substitute for Height in the Surface Area Formula We can substitute \( r \) in the surface area formula. First, we need to express \( h \) in terms of \( r \): From the surface area formula: \[ S = 2\pi rh + \pi r^2 \] Rearranging gives us: \[ h = \frac{S - \pi r^2}{2\pi r} \] ### Step 4: Substitute \( h \) in the Volume Formula Now, we substitute \( h \) back into the volume formula: \[ V = \pi r^2 \left( \frac{S - \pi r^2}{2\pi r} \right) \] This simplifies to: \[ V = \frac{r(S - \pi r^2)}{2} \] ### Step 5: Differentiate the Volume with Respect to \( r \) To find the maximum volume, we differentiate \( V \) with respect to \( r \): \[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{Sr}{2} - \frac{\pi r^3}{2} \right) \] This gives us: \[ \frac{dV}{dr} = \frac{S}{2} - \frac{3\pi r^2}{2} \] ### Step 6: Set the Derivative Equal to Zero To find the critical points, set the derivative equal to zero: \[ \frac{S}{2} - \frac{3\pi r^2}{2} = 0 \] This leads to: \[ S = 3\pi r^2 \] ### Step 7: Substitute Back to Find \( h \) Now, substitute \( S \) back into the equation for \( h \): Using \( S = 3\pi r^2 \): \[ h = \frac{3\pi r^2}{2\pi r} - \frac{r}{2} \] This simplifies to: \[ h = \frac{3r}{2} - \frac{r}{2} = r \] ### Step 8: Relate Diameter to Height Since \( h = r \), we can now express the diameter: \[ D = 2r = 2h \] Thus, we can conclude: \[ k = 2 \] ### Final Answer The value of \( k \) is \( 2 \). ---
Promotional Banner

Topper's Solved these Questions

  • APPLICATION OF DERIVATIVES

    PUNEET DOGRA|Exercise PREV YEAR QUESTIONS |85 Videos
  • 3-D GEOMETRY

    PUNEET DOGRA|Exercise PREV YEAR QUESTIONS|108 Videos
  • AREA BOUNDED BY CURVES

    PUNEET DOGRA|Exercise PREV YEAR QUESTIONS|39 Videos

Similar Questions

Explore conceptually related problems

A cylindrical jar without a lid has to be constructed using a given surface area of a metal sheet if the capacity of the jar times the height of the jar The value of k is

A parallel plate capacitor consists of two metal plates of area A and separation d . A slab of thickness t and electric constant K is inserted between the plates with its faces parallel to the plates and having the same surface area as that of the plates. Find the capacitance of the system. IF K=2 , for what value of t//d will the capacitance of the system be 3//2 times that of the air capacitor? Calculate the energy in the two cases and account for the energy change.

A conical vessel is to be prepared out of a circular sheet of metal of unit radius in order that the vessel has maximum value, the sectorial area that must be removed from the sheet is A_(1) and the area of the given sheet is A_(2) , then A_(2)/A_(1) is equal to

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.

In a toys manufacturing company, wooden parts are assembled and painted to prepare a toy. One specific toy is in the shape of a cone mounted on a cylinder. For the wood processing activity center, the wood is taken out of storage to be sawed, after which it undergoes rough polishing, then is cut, drilled and has holes punched in it. It is then fine polished using sandpaper. For the retail packaging and delivery activity center, the polished wood sub-parts are assembled together, then decorated using paint. The total height of the toy is 26 cm and the height of its conical part is 6 cm. The diameters of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. If its cylindrical part is to be painted yellow, the surface area need to be painted is

A parallel plate capacitor is to be constructed which can store q = 10 muC charge at V = 1000 volt. The minimum plate area of the capacitor is required to be A_(1) when space between the plates has air. If a dielectric of constant K = 3 is used between the plates, the minimum plate area required to make such a capacitor is A_(2) . The breakdown field for the dielectric is 8 times that of air. Find (A_(1))/(A_(2))

In a toys manufacturing company, wooden parts are assembled and painted to prepare a toy. One specific toy is in the shape of a cone mounted on a cylinder. For the wood processing activity center, the wood is taken out of storage to be sawed, after which it undergoes rough polishing, then is cut, drilled and has holes punched in it. It is then fine polished using sandpaper. For the retail packaging and delivery activity center, the polished wood sub-parts are assembled together, then decorated using paint. The total height of the toy is 26 cm and the height of its conical part is 6 cm. The diameters of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. If its conical part is to be painted green, the surface area need to be painted is

Two identical adiabatic containers of negligible heat capacity are connected by conducting rod of length L and cross sectional area A. Thermal conductivity of the rod is k and its curved cylindrical surface is well insulated from the surrounding. Heat capacity of the rod is also negligible. One container is filled with n moles of helium at temperature T_(1) and the other one is filled with equal number of moles of hydrogen at temperature T_(2) (lt T_(1)) . Calculate the time after which the temperature difference between two gases will becomes half the initial difference.

PUNEET DOGRA-APPLICATION OF DERIVATIVES-PREV YEAR QUESTIONS
  1. Match list-I with List-II and select the correct answer using the code...

    Text Solution

    |

  2. The maximum value of (ln x)/x is:

    Text Solution

    |

  3. A cylindrical jar without a lid has to be constructed using a given su...

    Text Solution

    |

  4. What is the maximum area of a triangle that can be inscribed in a circ...

    Text Solution

    |

  5. What is die length of the longest interval in which the function f(x) ...

    Text Solution

    |

  6. What is the maximum value of the function f(x) = 4 sin^(2)x + 1?

    Text Solution

    |

  7. Which one of the following statement is correct respect to kidney func...

    Text Solution

    |

  8. Consider the following function for the next two items that follow: ...

    Text Solution

    |

  9. Consider the following fur the next two items that follow: f(x) = {{...

    Text Solution

    |

  10. If x^(2) - px + 4 gt 0 for all real values of x. then which one of th...

    Text Solution

    |

  11. Let x and y be positive integers such that x is prime and y is composi...

    Text Solution

    |

  12. The number of integral values of a in [0,10) so that function, f(x)=x^...

    Text Solution

    |

  13. What is the number of points of local maxima of the function f(x)?

    Text Solution

    |

  14. Consider the function f(x) = (x^(2)-1)/(x^(2) + 1), where x in R Wha...

    Text Solution

    |

  15. Consider the function: f(x) = |x-1| + x^(2), where x in R Which on...

    Text Solution

    |

  16. Consider the function f(theta) =4(sin^(2)theta + cos^(4)theta) What...

    Text Solution

    |

  17. Consider the function f(theta) =4(sin^(2)theta + cos^(4)theta) What...

    Text Solution

    |

  18. Consider the function f(x) = (1/x)^(2x^(2)), where x gt 0 . At what...

    Text Solution

    |

  19. Consider the function f(x) = (1/x)^(2x^(2)), where x gt 0 . The mini...

    Text Solution

    |

  20. The function f(x)=2x^3-3x^2-12x+4 has

    Text Solution

    |