Answer:
the radius of the base is equal to [tex]x=2\sqrt[3]{\frac{105}{\pi } }[/tex]
And the height is equal to:
[tex]y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }[/tex]
Step-by-step explanation:
We write the volume function
[tex]f(x,y)=840=\pi x^{2} y[/tex] where x is the radius of the base and y is the height of the cylinder
[tex]y=\frac{840}{\pi x^{2} }[/tex]
The surface of a cylinder is given by
[tex]S(x)=\pi x^{2} +2\pi xy=\pi x^{2} +\frac{1680}{x}[/tex] on the interval from 0 to infinity
We now determine the critical values by differentiating and making the equation equal to 0
[tex]f'(x)=2\pi*x- \frac{1680}{x^{2}} =\frac{2\pi x^{3}-1680}{x^{2} } =0[/tex]
This equation have 2 complex solutions and one real solution
[tex]x=2\sqrt[3]{\frac{105}{\pi } }[/tex]
For x=0 and infinity the equation goes to infinity also so the radius of the base is equal to [tex]x=2\sqrt[3]{\frac{105}{\pi } }[/tex]
And the height is equal to:
[tex]y=\frac{840}{\pi (2\sqrt[3]{\frac{105}{\pi } })^{2} }[/tex]
[tex]y=\frac{840}{\pi(2\sqrt[3]{\frac{105}{\pi}})^{2} }[/tex]
