19. minimum - surface - area box of all boxes with a square base and a volume of 8m³, which one has the…

Explanation:

Step1: Define variables

Let the side length of the square base be (x) (in meters) and the height of the box be (h) (in meters). The volume (V=x^{2}h), and since (V = 8), we have (h=\frac{8}{x^{2}}). The surface - area formula (S=2x^{2}+4xh).

Step2: Substitute (h) into the surface - area formula

Substitute (h=\frac{8}{x^{2}}) into (S): (S(x)=2x^{2}+4x\cdot\frac{8}{x^{2}}=2x^{2}+\frac{32}{x}), where (x>0).

Step3: Find the derivative of (S(x))

Using the power rule, (S^\prime(x)=4x-\frac{32}{x^{2}}).

Step4: Set the derivative equal to zero and solve for (x)

Set (S^\prime(x) = 0): [ \begin{align*} 4x-\frac{32}{x^{2}}&=0\ 4x&=\frac{32}{x^{2}}\ x^{3}& = 8\ x&=2 \end{align*} ]

Step5: Find the second - derivative of (S(x))

(S^{\prime\prime}(x)=4+\frac{64}{x^{3}}). When (x = 2), (S^{\prime\prime}(2)=4+\frac{64}{8}=4 + 8=12>0). So (S(x)) has a minimum at (x = 2).

Step6: Find the value of (h)

When (x = 2), (h=\frac{8}{2^{2}}=2).

Answer:

The box with dimensions (2\space m\times2\space m\times2\space m) (a cube) has the minimum surface area.