Please show step by step of working out the value of r for which is A is aminimum and calculate the minimum surface area of the container.
The total surface area, of each container is modelled by function A= πr^2+1000/r.

(remember to use the derivative to show you have found the minimum)​

Please show step by step of working out the value of r for which is A is aminimum and calculate the minimum surface area of the container The total surface area class=

Respuesta :

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.

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