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Question: A cylindrical jar without a lid has to be constructed using a given surface area of a metal sheet. I...

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 is to be 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

Explanation

Solution

In this question remember to identify the surface area of the metal sheet which is mentioned constant in the given information also remember to use the application of derivative as for maximum capacity of the jar dVdD\dfrac{{dV}}{{dD}} must be equal to 0, using this information will help you to approach the solution of the question.

Complete step-by-step answer :
According to the given information we know that cylindrical jar where diameter is equal to the k times the height of jar also the cylindrical jar is made of metal sheet whose surface area has given
And we know that cylinder contains two circular bases and lateral surface area
Therefore, surface area of metal sheet = lateral surface area + base of cylinder
Also, we know that formula of lateral surface and circular surface area is given as; lateral surface area=πDH = \pi DH, base of cylinder=πD24 = \pi \dfrac{{{D^2}}}{4}
So, S=πDH+πD24S = \pi DH + \pi \dfrac{{{D^2}}}{4}
As we require value of H for volume of the cylinder
SπD24=πDHS - \pi \dfrac{{{D^2}}}{4} = \pi DH
Or1πD(SπD24)=H\dfrac{1}{{\pi D}}\left( {S - \dfrac{{\pi {D^2}}}{4}} \right) = H
OrH=SπDD4H = \dfrac{S}{{\pi D}} - \dfrac{D}{4} (equation 1)
As, we know that volume of cylinder is given as; V=πD24HV = \pi \dfrac{{{D^2}}}{4}H here D is the diameter of the circular bases of the cylinder and H is the height
Substituting the value of H from equation 1 in the formula of cylinder we get
V=πD24(SπDD4)V = \pi \dfrac{{{D^2}}}{4}\left( {\dfrac{S}{{\pi D}} - \dfrac{D}{4}} \right)
OrV=DS4πD316V = \dfrac{{DS}}{4} - \dfrac{{\pi {D^3}}}{{16}} (equation 2)
We know that for maximum volumedVdD\dfrac{{dV}}{{dD}} must be equal to zero i.e. dVdD=0\dfrac{{dV}}{{dD}} = 0
Now differentiating equation 2 with respect to D
dVdD=ddDDS4ddDπD316\dfrac{{dV}}{{dD}} = \dfrac{d}{{dD}}\dfrac{{DS}}{4} - \dfrac{d}{{dD}}\dfrac{{\pi {D^3}}}{{16}}
Since, we know that ddx(x)=1\dfrac{d}{{dx}}\left( x \right) = 1 and ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}
Therefore, dVdD=S43πD3116\dfrac{{dV}}{{dD}} = \dfrac{S}{4} - \dfrac{{3\pi {D^{3 - 1}}}}{{16}}
\Rightarrow $$$$\dfrac{{dV}}{{dD}} = \dfrac{S}{4} - \dfrac{{3\pi {D^2}}}{{16}}
Since, for maximum volume dVdD=0\dfrac{{dV}}{{dD}} = 0
Therefore, S43πD216=0\dfrac{S}{4} - \dfrac{{3\pi {D^2}}}{{16}} = 0
\Rightarrow $$$\dfrac{{4S - 3\pi {D^2}}}{{16}} = 0$ \Rightarrow 4S - 3\pi {D^2} = 0$ Or$S = \dfrac{{3\pi {D^2}}}{4}$ Substituting the value of “S” in equation 1 we get $H = \dfrac{{3\pi {D^2}}}{{4\pi D}} - \dfrac{D}{4}$ $$ \Rightarrow H = \dfrac{{3D}}{4} - \dfrac{D}{4} $$ \Rightarrow $$$H = \dfrac{{3D - D}}{4}
OrH=D2H = \dfrac{D}{2}
Hence, the diameter D is 2 times the height
Hence, B is the correct option.

Note : Whenever we face these types of question the key concept is that the we have to know the formula of the surface area of cylinder i.e. surface area of metal sheet = lateral surface area + base of cylinder and here in this question we find the surface area of metal sheet then we find the volume of cylinder and to get maximum volume dVdD=0\dfrac{{dV}}{{dD}} = 0 and after getting H and S we will get the relationship between height and diameter.