Erfana Beevee Binte Hassan Shariff's answer to Ryan's Secondary 4 A Maths Singapore question.
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Part 2/2 of solution.
First, express the total surface area of figure in terms of h and r and equate to 24pi.
Then express h in terms of r since the volume to be shown is in terms of r.
Then express volume of figure in terms of h and r and sub the expression of h into it.
Then differentiate w.r.t r. Equate dV/dr=0 and you will get 2 values.
In fact, r cannot take negative value. So the negative value can be rejected. So the volume is a maximum when r takes the positive value.
Hope this helps :)
First, express the total surface area of figure in terms of h and r and equate to 24pi.
Then express h in terms of r since the volume to be shown is in terms of r.
Then express volume of figure in terms of h and r and sub the expression of h into it.
Then differentiate w.r.t r. Equate dV/dr=0 and you will get 2 values.
In fact, r cannot take negative value. So the negative value can be rejected. So the volume is a maximum when r takes the positive value.
Hope this helps :)
Date Posted:
7 months ago
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