# [SOLVED]Minimizing the cost of a boiler

#### leprofece

##### Member
it is necessary to manufacture a boiler, composed of a cylinder and two funds hemispherical, with walls of uniform thickness, and for a given volume V. Calculate the dimensions of the boiler which can be built with the least amount of material. r = cubic sqrt of ( 3v/4pi)

V = 2pir2H + 4/3pir3
Cylinder and sphere i cant find out how to link the equations

#### MarkFL

Staff member
Re: max and min 277

You 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?

#### leprofece

##### Member
Re: max and min 277

You 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?
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 r2 +2 pirh

#### MarkFL

Staff member
Re: max and min 277

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 r2 +2 pirh
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.

#### leprofece

##### Member
Re: max and min 277

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.
Lets see I almost got the answer i got cubic root of 3v/16 pi

from V equation I soplved and got H = V/2pir2-2R/3
Solving in a i got 8pir2/3+ V/R
I derived
16/3 pir3- V
And I equaled to 0
and i solved for R
I got cubic root of 3v/16 pi
IS IT THE BOOK WRONG????

#### MarkFL

Staff member
Re: max and min 277

No, 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.

#### leprofece

##### Member
Re: max and min 277

No, 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.
i reviewed and i realize i used the same equation of mi first post
so i cant find out the mistake

Last edited:

#### MarkFL

Staff member
Re: max and min 277

The 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:

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

#### leprofece

##### Member
Re: max and min 277

The 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:

$$\displaystyle V=\pi r^2h+\frac{4}{3}\pi r^3$$
Yeah I used that maybe the mistake is in the calculations but i could not find it

#### MarkFL

Staff member
Re: max and min 277

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.

#### leprofece

##### Member
Re: max and min 277

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.
Lets see I almost got the answer i got cubic root of (3v/16 pi)

from V equation I solved and got H = V/(2pir2)-2R/3
Solving i got 8pir2/3+ V/R
I derived
16/(3 pir3)- V
And I equaled to 0
and i solved for R
I got cubic root of 3v/16 pi

#### MarkFL

Staff member
Re: max and min 277

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.