How do you solve the optimization problem(calculus)?
tip: solve for h in the volume of a cone, then put that where h is for the Area of the cone.
Answer:
First you need relations for the volume and area of a cone. The volume is well known:
v = π r^2 h/3
To get the area, consider the pattern when a cone is unwrapped. It is a segment of a circle of radius b. Where:
b = √(h^2 + r^2)
The circumference of the pattern is the cicumference of the assembled cone (2 π r). The circumference of a complete circle is 2 π b so the fraction of the circle is the circumference ratio: 2π r / (2π b) = r/b so the area of the cone is:
A = π r b = π r √(h^2 + r^2)
To find the maximum volume for a given area, substitute the area realtion into the volume relation. To do this, invert the area relation:
A = π r √(h^2 + r^2)
h = √((A/(π r))^2 - r^2)
Substituting into the volume:
v = π r^2 h/3 = π r^2 √((A/(π r))^2 - r^2)/3
Simplify a little:
v = π r^2 √((A/(π r))^2 - r^2)/3 = √(A^2 r^2 - π^2 r^6 )/3
Determine dv/dr and set it to zero to find the r that gives maximum volume:
dv/dr = (1/2)((A^2 r^2 - π^2 r^6 )^(-1/2) * (2rA^2 - 6π^2 r^5)/3
To equal zero, only the variable term in the numberator need be zero:
2rA^2 - 6π^2 r^5 = 0
A^2 = (6π^2 r^5)/(2r) = 3 π^2 r^4
The area formula from above: A = π r √(h^2 + r^2) so:
A^2 = π^2 r^2 (h^2 + r^2)
Equating the two:
3 π^2 r^4 = π^2 r^2 (h^2 + r^2)
3 r^4 = h^2 r^2 + r^4
h = √(2) r
So the height is the square root of 2 times the radius, not the square of 2 times the radius as stated in the question.
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