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- #1

V = 2pir

^{2}H + 4/3pir

^{3}

Cylinder and sphere i cant find out how to link the equations

- Thread starter leprofece
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- Thread starter
- #1

V = 2pir

Cylinder and sphere i cant find out how to link the equations

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- #3

I dont know the amount of material The problem says what i writeYou have stated the constraint, but you need also the objective function, that is, the function you wish to optimize, which is the amount of materila needed to construct the boiler. What measure of the boiler will we need?

Maybe is it the area??

so it would be the area of the sphere 4 pi r

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Yes, good...the surface area is a measure of the amount of material needed. So, I recommend solving the constraint for $H$ and the substituting into the objective function to obtain a function in one variable, and then minimizing this function. Can you state the objective function as a function of $r$? And recall $V$ is a constant.I dont know the amount of material The problem says what i write

Maybe is it the area??

so it would be the area of the sphere 4 pi r^{2}+2 pirh

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Lets see I almost got the answer i got cubic root of 3v/16 piYes, good...the surface area is a measure of the amount of material needed. So, I recommend solving the constraint for $H$ and the substituting into the objective function to obtain a function in one variable, and then minimizing this function. Can you state the objective function as a function of $r$? And recall $V$ is a constant.

from V equation I soplved and got H = V/2pir

Solving in a i got 8pir

I derived

16/3 pir

And I equaled to 0

and i solved for R

I got cubic root of 3v/16 pi

IS IT THE BOOK WRONG????

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i reviewed and i realize i used the same equation of mi first postNo, your book is correct. I didn't notice before, but you have stated the constraint incorrectly. Look again at your first post regarding the volume of the boiler.

so i cant find out the mistake

Last edited:

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Yeah I used that maybe the mistake is in the calculations but i could not find itThe volume of the boiler is that of a cylinder and a sphere. You have twice the volume of a cylinder and then a sphere. So you want:

\(\displaystyle V=\pi r^2h+\frac{4}{3}\pi r^3\)

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Lets see I almost got the answer i got cubic root of (3v/16 pi)In the work you posted, it appears that you used the erroneous constraint. However, it was hard to read your work because you did not use bracketing symbols to clearly express rational terms.

from V equation I solved and got H = V/(2pir

Solving i got 8pir

I derived

16/(3 pir

And I equaled to 0

and i solved for R

I got cubic root of 3v/16 pi

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- #12

I can't really tell what you are doing. Ordinarily I would not do this, but I feel we are not making progress, so I am going to show you how to work this problem...

Okay, we have the objective function, the surface area of the boiler:

\(\displaystyle S(h,r)=2\pi rh+4\pi r^2\)

and this is subject to the constraint:

\(\displaystyle V=\pi r^2h+\frac{4}{3}\pi r^3\)

Now, if we solve the constraint for $h$, we obtain:

\(\displaystyle h=\frac{3V-4\pi r^3}{3\pi r^2}\)

Now, substituting this into the objective function, we obtain:

\(\displaystyle S(r)=2\pi r\left(\frac{3V-4\pi r^3}{3\pi r^2} \right)+4\pi r^2\)

Simplify:

\(\displaystyle S(r)=2\left(\frac{3V-4\pi r^3}{3r} \right)+4\pi r^2\)

\(\displaystyle S(r)=2Vr^{-1}+\frac{4}{3}\pi r^2\)

Now, differentiate with respect to $r$ and equate the result to zero to obtain the critical value(s):

\(\displaystyle S'(r)=-2Vr^{-2}+\frac{8}{3}\pi r=\frac{2\left(4\pi r^3-3V \right)}{3r^2}=0\)

This implies (for $0<r$):

\(\displaystyle 4\pi r^3-3V=0\)

Solving for $r$, we find:

\(\displaystyle r=\sqrt[3]{\frac{3V}{4\pi}}\)

We can see that for any positive $r$, we have $S''(r)>0$, thus we know this critical value is at a minimum.