4. a can is made from a sheet of metal of 600 cm². determine the dimensions of a can that will maximize…

Explanation:

Step1: Set up surface - area and volume formulas

The surface - area formula of a cylinder is $S = 2\pi r^{2}+2\pi rh$, and we know $S = 600$, so $2\pi r^{2}+2\pi rh=600$. Solving for $h$, we get $h=\frac{600 - 2\pi r^{2}}{2\pi r}=\frac{300}{\pi r}-r$. The volume formula of a cylinder is $V=\pi r^{2}h$.

Step2: Substitute $h$ into the volume formula

Substitute $h=\frac{300}{\pi r}-r$ into $V=\pi r^{2}h$, we have $V=\pi r^{2}(\frac{300}{\pi r}-r)=300r-\pi r^{3}$.

Step3: Differentiate the volume function

Differentiate $V(r)=300r - \pi r^{3}$ with respect to $r$. Using the power rule $\frac{d}{dr}(x^{n})=nx^{n - 1}$, we get $V^\prime(r)=300 - 3\pi r^{2}$.

Step4: Find the critical points

Set $V^\prime(r)=0$, so $300 - 3\pi r^{2}=0$. Then $3\pi r^{2}=300$, and $r^{2}=\frac{100}{\pi}$, so $r=\sqrt{\frac{100}{\pi}}=\frac{10}{\sqrt{\pi}}\text{ cm}$.

Step5: Find the height $h$

Substitute $r = \frac{10}{\sqrt{\pi}}$ into the formula for $h$: $h=\frac{300}{\pi\times\frac{10}{\sqrt{\pi}}}-\frac{10}{\sqrt{\pi}}=\frac{30}{\sqrt{\pi}}-\frac{10}{\sqrt{\pi}}=\frac{20}{\sqrt{\pi}}\text{ cm}$.

Answer:

The radius of the can is $r = \frac{10}{\sqrt{\pi}}\text{ cm}$ and the height is $h=\frac{20}{\sqrt{\pi}}\text{ cm}$.