As part of an Art Project, a student designed a lantern in the shape of an octahedron as shown in the diagram. The frame of the lantern consists of twelve straight pieces of wire, joined at their ends to form two identical right pyramids with a square base of sides $4x\text{ m}$ and slant edges of $y\text{ m}$. The wire frame is covered tightly with coloured paper. The wire cost $8 per metre and the coloured paper cost $15 per square metre. The student is required to include the budget for his lantern in this project. Given that the total length of the wire used is $3.2\text{ m}$, use differentiation to find the maximum amount for his budget, giving your answer to the nearest dollar.

1. Express the slant edge $y$ in terms of $x$ using the total length of the wire.

The wire frame consists of the square base (4 edges of length $4x$) and the slant edges of the two pyramids (8 edges of length $y$).
Total length of wire $= 4(4x) + 8y = 16x + 8y$
Given that the total length of the wire used is $3.2\text{ m}$:
$16x + 8y = 3.2$
$8y = 3.2 - 16x$
$y = 0.4 - 2x$

2. Find an expression for the total surface area of the lantern in terms of $x$ and $y$.

The surface of the lantern is formed by 8 identical isosceles triangular faces.
For one triangular face with base $4x$ and two slant edges $y$, the height $h$ is given by Pythagoras' theorem:
$h = \sqrt{y^2 - (2x)^2} = \sqrt{y^2 - 4x^2}$
Area of one triangular face $= \frac{1}{2}(4x)h = 2x\sqrt{y^2 - 4x^2}$
Total surface area $A = 8 \times 2x\sqrt{y^2 - 4x^2} = 16x\sqrt{y^2 - 4x^2}$

3. Substitute $y$ to express the total budget $C$ entirely in terms of $x$.

Substitute $y = 0.4 - 2x$ into $A$:
$A = 16x\sqrt{(0.4 - 2x)^2 - 4x^2}$
$A = 16x\sqrt{0.16 - 1.6x + 4x^2 - 4x^2} = 16x\sqrt{0.16 - 1.6x}$
The total budget $C$ is the sum of the wire cost and the paper cost.
Cost of wire $= 8 \times 3.2 = 25.6$
Cost of paper $= 15 \times A = 15 \times 16x\sqrt{0.16 - 1.6x} = 240x\sqrt{0.16 - 1.6x}$
$C = 25.6 + 240x\sqrt{0.16 - 1.6x}$

4. Differentiate $C$ with respect to $x$ and equate to $0$ to find the stationary point.

Using the product rule and chain rule to find $\frac{\mathrm{d}C}{\mathrm{d}x}$:
$\frac{\mathrm{d}C}{\mathrm{d}x} = 240 \left[ 1 \cdot \sqrt{0.16 - 1.6x} + x \cdot \frac{1}{2\sqrt{0.16 - 1.6x}}(-1.6) \right]$
$\frac{\mathrm{d}C}{\mathrm{d}x} = 240 \left[ \sqrt{0.16 - 1.6x} - \frac{0.8x}{\sqrt{0.16 - 1.6x}} \right]$
To maximize the budget, let $\frac{\mathrm{d}C}{\mathrm{d}x} = 0$:
$\sqrt{0.16 - 1.6x} - \frac{0.8x}{\sqrt{0.16 - 1.6x}} = 0$
Multiply through by $\sqrt{0.16 - 1.6x}$:
$0.16 - 1.6x - 0.8x = 0$
$0.16 - 2.4x = 0$
$x = \frac{0.16}{2.4} = \frac{1}{15}$

5. Verify that this value of $x$ gives a maximum budget using the First Derivative Test.

Testing the sign of $\frac{\mathrm{d}C}{\mathrm{d}x}$ around $x = \frac{1}{15} \approx 0.0667$:

• When $x = 0.06$, $\frac{\mathrm{d}C}{\mathrm{d}x} \propto 0.16 - 2.4(0.06) = 0.016 > 0$
• When $x = 0.07$, $\frac{\mathrm{d}C}{\mathrm{d}x} \propto 0.16 - 2.4(0.07) = -0.008 < 0$
  Since the gradient changes from positive to negative, $x = \frac{1}{15}$ yields a maximum point.

6. Calculate the maximum budget and round it to the nearest dollar.

Substitute $x = \frac{1}{15}$ into the budget equation:
$C = 25.6 + 240\left(\frac{1}{15}\right)\sqrt{0.16 - 1.6\left(\frac{1}{15}\right)}$
$C = 25.6 + 16\sqrt{\frac{4}{25} - \frac{16}{150}}$
$C = 25.6 + 16\sqrt{\frac{24 - 16}{150}} = 25.6 + 16\sqrt{\frac{8}{150}}$
$C = 25.6 + 16\sqrt{\frac{4}{75}} = 25.6 + \frac{32}{5\sqrt{3}}$
$C \approx 25.6 + 3.6950 = 29.295$
Rounding to the nearest dollar, the maximum budget is $\char36{}29$.