Minimizing tin cost with volume constraint and doubled metal price

[danago]

"A company is to produce tins of volume 535cm$^3$. The cost of the metal per square meter for the base and top is twice the cost of the walls. What base radius and tin height should be used for an optimal price?"

First i wrote equations for the surface area, volume, and cost.

$$\displaylines{<br /> A_{Side} = 2\pi rh \cr <br /> A_{Ends} = 2\pi r^2 \cr <br /> C_{Side} = x(2\pi rh) \cr <br /> C_{Ends} = 2x(2\pi r^2 ) \cr <br /> C = 2x(2\pi r^2 + \pi rh) \cr}<br /> 535 = \pi r^2 h\therefore h = \frac{{535}}{{\pi r^2 }}$$

I then rewrote my cost equation, replacing 'h' with an expression in terms of 'r'.
$$C = 2x(2\pi r^2 + \frac{{535}}{r})$$

So now i have an equation for cost, in terms of x (the cost of the metal for the walls of the tin), and r (the radius of the base/top). Have i gone about it the right way? If so, what should i do next? can i just cancel the x off?

Thanks,
Dan.

 

[arunbg]

Yes your initial working is absolutely correct.
All you have to do now is to minimise the final cost function C using calculus . Remember that x is a constant here and so the only variable is r.

 

[danago]

Yes your initial working is absolutely correct.
All you have to do now is to minimise the final cost function C using calculus . Remember that x is a constant here and so the only variable is r.

ok. So using the product rule i get:
$$C' = 2x(4\pi r - \frac{{535}}{{r^2 }})$$

But I've still got two variables.

 

[arunbg]

You have to actually set $$\frac{dC}{dr}=0$$ for minimum.
Also remember that x is a constant and can be eliminated.
Now find r.

 

[danago]

ok i see what do do now. If i had actually taken the time to continue with my working instead of automatically think it was wrong, i would have got it :)

So since
$$2x(4\pi r - \frac{{535}}{{r^2 }}) = 0$$

I divide everything by 2x, which ends up cancelling the x anyway :) And the answer would be r=3.5, and h=14 :)

Thanks a lot for the help :P

 

[arunbg]

You're welcome.:D