Answer:
The minimum surface area of the container is 276.791 square units.
Step-by-step explanation:
Let be [tex]A(r) = \pi\cdot r^{2} + \frac{1000}{r}[/tex], [tex]\forall \,r \in \mathbb{R}[/tex], [tex]r \geq 0[/tex]. The first and second derivatives of such function are, respectively:
First derivative
[tex]A'(r) = 2\cdot \pi \cdot r -\frac{1000}{r^{2}}[/tex]
Second derivative
[tex]A''(r) = 2\cdot \pi +\frac{2000}{r^{3}}[/tex]
The critical values of [tex]r[/tex] are determined by equalizing first derivative to zero and solving it: (First Derivative Test)
[tex]2\cdot \pi \cdot r -\frac{1000}{r^{2}} = 0[/tex]
[tex]2\cdot \pi \cdot r^{3} - 1000 = 0[/tex]
[tex]r = \sqrt[3]{\frac{1000}{2\pi} }[/tex]
[tex]r \approx 5.419[/tex] (since radius is a positive variable)
To determine if critical value leads to an absolute minimum, this input must be checked in the second derivative expression: ([tex]r \approx 5.419[/tex])
[tex]A''(5.419) = 2 + \frac{2000}{5.419^{3}}[/tex]
[tex]A''(5.419) = 14.568[/tex]
The critical value leads to an absolute minimum, since value of the second derivative is positive.
Finally, the minimum surface area of the container is:
[tex]A(5.419) = \pi\cdot (5.419)^{2} + \frac{1000}{5.419}[/tex]
[tex]A(5.419) \approx 276.791[/tex]
The minimum surface area of the container is 276.791 square units.