Let's go through each question and solve them step-by-step.

Question 1

An open box is formed from a card. The volume of the box is given by: $V = 4x^3 - 78x^2 + 360x \, \text{cm}^3$

1. Part (i): Show that this is the volume of the box.

   • We don't have the exact details of the folding, but if this is provided as the volume, it’s taken as given here.

2. Part (ii): Find the value of $x$ that maximizes the volume.

   • To find the maximum volume, we differentiate $V$ with respect to $x$ and set the derivative equal to zero.

$V = 4x^3 - 78x^2 + 360x$ $\frac{dV}{dx} = 12x^2 - 156x + 360$ Setting $\frac{dV}{dx} = 0$ to find the critical points: $12x^2 - 156x + 360 = 0$ Solve this quadratic equation for $x$ to find the value that maximizes the volume.

Question 2

A cylinder is cut from a solid sphere of radius 3 cm. The height of the cylinder is $2h$.

1. Part (i): Find the radius of the cylinder in terms of $h$.

   • The cylinder fits within a sphere with radius 3 cm. Using the Pythagorean theorem in the circular cross-section, we find: $r^2 + h^2 = 3^2$ $r = \sqrt{9 - h^2}$

2. Part (ii): Show that the volume of the cylinder is $V = 2\pi h (9 - h^2)$.

   • The volume $V$ of a cylinder is given by $V = \pi r^2 \cdot \text{height}$.
   • Substitute $r = \sqrt{9 - h^2}$ and height $2h$: $V = \pi (\sqrt{9 - h^2})^2 \cdot 2h = 2\pi h (9 - h^2)$

3. Part (iii): Find the maximum volume of the cylinder as $h$ varies.

   • Differentiate $V = 2\pi h (9 - h^2)$ with respect to $h$ and set $\frac{dV}{dh} = 0$ to find the value of $h$ that maximizes the volume.

Question 3

A cylindrical oil tank with radius $r$ and height $h$ has a combined height and radius sum of 24 m.

1. To maximize the volume, express $h$ in terms of $r$ as $h = 24 - r$, then write the volume $V = \pi r^2 h$ in terms of $r$ alone: $V = \pi r^2 (24 - r)$ Then, differentiate with respect to $r$ and set $\frac{dV}{dr} = 0$ to find the value of $r$ that maximizes the volume.

Question 4

A cylindrical can with height $h$ and radius $r$ has a volume of 2 liters.

1. Part (i): Express $h$ in terms of $r$.

   • Given $V = \pi r^2 h = 2$: $h = \frac{2}{\pi r^2}$

2. Part (ii): Find an expression for the surface area in terms of $r$ only.

   • Surface area $A = 2\pi r h + 2\pi r^2$: $A = 2\pi r \cdot \frac{2}{\pi r^2} + 2\pi r^2 = \frac{4}{r} + 2\pi r^2$

3. Part (iii): Find $r$ that minimizes the surface area.

   • Differentiate $A$ with respect to $r$ and set $\frac{dA}{dr} = 0$ to find the value of $r$ that minimizes the surface area.

Question 5

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Would you like detailed solutions for each step? Here are 5 related questions that expand on the concepts:

1. How do we find critical points to determine maximum or minimum values?
2. Why is differentiation useful in optimization problems?
3. How do we handle constraints in optimization, like fixed volume or combined dimensions?
4. What’s the geometric interpretation of maximizing volume in a constrained shape?
5. How do you apply the second derivative test to confirm maximum/minimum?

Tip: Always double-check units and constraints in optimization problems—they’re crucial for setting up correct equations!