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To solve, apply the formula of volume of a cylinder.
Plug-in the given volume of the cylindrical tank.
Then, divide both sides by `pir^2` to isolate the h.
`512/r^2=h` (Let this be EQ1.)
Since the material of the cylindrical tank (with open top) must be minimized, set-up the equation of its area.
Area = Area of the bottom + Area of side
To express the right side as one variable, substitute EQ1.
To solve for r that would minimize the area, take the derivative of A.
`A=pir^2 + 1024pir^(-1)`
Then, set A' equal to zero.
To simplify the equation, divide both sides by 2pi.
`0=r - 512/r^2`
And multiply both sides by r^2.
`r^2*0=(r - 512/r^2)*r^2`
Then, isolate r. So, add both sides by 512.
And take the cube root of both sides.
Next, solve for h. To do so, plug-in the value of r to EQ1.
Hence, to minimize the area of the cylindrical tank, its radius and height should be both 8 meters.
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