"A company is to produce tins of volume 535cm[itex]^3[/itex]. 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?"(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]

\displaylines{

A_{Side} = 2\pi rh \cr

A_{Ends} = 2\pi r^2 \cr

C_{Side} = x(2\pi rh) \cr

C_{Ends} = 2x(2\pi r^2 ) \cr

C = 2x(2\pi r^2 + \pi rh) \cr}

535 = \pi r^2 h\therefore h = \frac{{535}}{{\pi r^2 }}

[/tex]

I then rewrote my cost equation, replacing 'h' with an expression in terms of 'r'.

[tex]

C = 2x(2\pi r^2 + \frac{{535}}{r})

[/tex]

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.

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# Homework Help: Optimal Cost

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