SOLUTION: A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of m

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Question 1075057: A cylindrical biscuit tin has a close-fitting lid which overlaps the tin by 1cm. The radii of the tin and the lid are both x cm. The tin and the lid are made from a thin sheet of metal of area 80π square cm and there is no wastage. The volume of the tin is V cubic cm. Show that V=π (40x-x^2-x^3). Use differentiation to find the positive value of x for which V is stationary.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
h= height of the tin in cm.
The surface area of the tin and lid (in square cm) will be
2pi%2Ax%5E2%2B2pi%2Ax%2A%28h%2B1%29=80pi
Dividing everything by 2pi we get
x%5E2%2B%28h%2B1%29x=40 ---> h%2B1=%2840-x%5E2%29%2Fx ---> h%2B1=40%2Fx-x%5E2%2Fx ---> h%2B1=40%2Fx-x ---> h=40%2Fx-1-x

The volume of the tin as a function of x is
V%28x%29=pi%2Ax%5E2%2Ah
Substituting the expression previously found for h
V%28x%29=pi%2Ax%5E2%2A%2840%2Fx-1-x%29
V%28x%29=pi%2840x-x%5E2-x%5E3%29

Obviously we want to make the shape with the largest possible volume,
so we find the radius x that will give us maximum volume.
For that we calculate the derivative:
dV%2Fdx=pi%2840-2x-3x%5E2%29
The quadratic polynomial 40-2x-3x%5E2=%2810-3x%29%28x%2B4%29
has zeros at x=10%2F3 and x=-4 ,
changing from positive to negative at highlight%28x=10%2F3%29 .
That is the x value that will yield the maximum volume,
so the radius of the tin should be highlight%2810+%2F+3%29 cm ,
or about highlight%283.3cm%29 .

NOTE: The other zero of the derivative, x=-4 , is a minimum of the function,
but a negative value for the radius has no practical meaning.
The problem tells you to look for the positive value,
just in case you do not understand the problem, but can follow problem-solving recipes.