What are the dimensions of the most economical cylindrical can?

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To determine the dimensions of the most economical cylindrical can holding 20 m³, the optimal height (h) is 5 meters and the radius (r) is 2 meters. The volume equation is V = πr²h, which equals 20π when substituting the values. The surface area cost is calculated using the equations for the top, bottom, and side of the can, leading to a total cost function. Differentiation is then applied to find the minimum surface area, confirming the optimal dimensions. This approach effectively balances material costs for the can's construction.
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A cylindrical can is to hold 20 m3. The material for the top and bottom costs
€10/m^2 and material for the side costs €8/m^2. Find the radius r and height h of
the most economical can.

the answer is h= 5 and j=2

i have found the eqn such as v=pi j^2 h =20 pi

and 2pi j^2 + pi j h = 18...then i don't know how to do...pls help
 
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Have you learned yet how to use differentiation to find maxima and minima of a function?
 
ya...but is my eqn correct??
 
What's "j"? Where's "r" in the equations?
 
j = radius
 
Well, anyway, I'd solve it this way:

-- write an equation 1 for V = f(r,h)

-- write an equation 2 for A = f(r,h)

-- rearrange equation 1 so you can substitute it into equation 2 to eliminate r in equation 2.

-- differentiate the result with respect to h, and set the result = 0

-- solve for the height h that minimizes the area, and then use equation 1 to solve for r.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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