1. Define dimensions of the rectangle: Let the length be \( 2r \) and the width be \( h \), where \( 2r \) is the circumference of the cylinder base and \( h \) is its height. The perimeter is: \[ 2r + 2h = 300 \quad \Rightarrow \quad r + h = 150 \quad \Rightarrow \quad h = 150 - r. \] 2. Volume of the cylinder: The volume of the cylinder is: \[ V = \pi r^2 h = \pi r^2 (150 - r). \] 3. Maximize \( V \): Differentiate \( V \) with respect to \( r \): \[ \frac{dV}{dr} = \pi \left[ 2r(150 - r) - r^2 \right] = \pi (300r - 3r^2). \] Set \( \frac{dV}{dr} = 0 \): \[ 300r - 3r^2 = 0 \quad \Rightarrow \quad 3r(100 - r) = 0. \] Thus, \( r = 0 \) or \( r = 100 \). Discard \( r = 0 \) since it gives no volume.
4. Second derivative test: \[ \frac{d^2V}{dr^2} = \pi (300 - 6r). \] At \( r = 100 \): \[ \frac{d^2V}{dr^2} = \pi (300 - 600) = -300\pi<0. \] Hence, \( V \) is maximum at \( r = 100 \).
5. Find \( h \): \[ h = 150 - r = 150 - 100 = 50. \]
Final Answer: The dimensions of the rectangular sheet are \( 2r = 200 \, {cm} \) and \( h = 50 \, {cm} \).
If a random variable X has the following probability distribution values:
X | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
---|---|---|---|---|---|---|---|---|
P(X) | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 | 1/12 |
Then P(X ≥ 6) has the value: