1. **Problem statement:**
Kailash manufactures a cuboid container with a square base of side length $l$ cm and height $d$ cm, where $d$ is three times $l$. The container holds 375 cm³ of drink. We need to find expressions for $d$, values of $l$ and $d$, the total surface area, then consider a cylindrical container with the same volume and minimize its surface area.
2. **Expression for $d$ in terms of $l$:**
Given height $d$ is three times the base length $l$, so
$$d = 3l$$
3. **Find $l$ and $d$ given volume 375 cm³:**
Volume of cuboid $V = l^2 \times d$
Substitute $d = 3l$:
$$V = l^2 \times 3l = 3l^3$$
Given $V = 375$, so
$$3l^3 = 375$$
Divide both sides by 3:
$$\cancel{3}l^3 = \cancel{3}125$$
$$l^3 = 125$$
Take cube root:
$$l = \sqrt[3]{125} = 5$$
Then height:
$$d = 3l = 3 \times 5 = 15$$
4. **Calculate total external surface area of the cuboid:**
Surface area $S = 2 \times$ (area of base + area of sides)
Base is square: area $= l^2 = 5^2 = 25$
Sides: 4 rectangles each $l \times d = 5 \times 15 = 75$
Total side area $= 4 \times 75 = 300$
Total surface area:
$$S = 2(25 + 300) = 2 \times 325 = 650$$
5. **Expression for height $h$ of cylinder in terms of radius $r$:**
Volume of cylinder $V = \pi r^2 h$
Given $V = 375$, solve for $h$:
$$h = \frac{375}{\pi r^2}$$
6. **Show that total surface area $A = 2\pi r^2 + \frac{750}{r}$:**
Surface area of cylinder:
$$A = 2\pi r^2 + 2\pi r h$$
Substitute $h$:
$$A = 2\pi r^2 + 2\pi r \times \frac{375}{\pi r^2} = 2\pi r^2 + \frac{750}{r}$$
7. **Find derivative $\frac{dA}{dr}$:**
$$\frac{dA}{dr} = \frac{d}{dr} \left(2\pi r^2 + \frac{750}{r}\right) = 4\pi r - \frac{750}{r^2}$$
8. **Find $r$ that minimizes $A$:**
Set derivative to zero:
$$4\pi r - \frac{750}{r^2} = 0$$
Multiply both sides by $r^2$:
$$4\pi r^3 = 750$$
Divide both sides by $4\pi$:
$$r^3 = \frac{750}{4\pi} = \frac{375}{2\pi}$$
Take cube root:
$$r = \sqrt[3]{\frac{375}{2\pi}}$$
9. **Find minimum value of $A$:**
Substitute $r$ back into $A$:
$$A = 2\pi r^2 + \frac{750}{r}$$
Use $r^3 = \frac{375}{2\pi}$ to simplify if needed.
10. **Compare total material needed including additional surface area:**
Cuboid material:
$$650 + 0.1 \times 650 = 650 \times 1.1 = 715$$
Cylinder material:
$$A_{min} + 0.25 \times A_{min} = A_{min} \times 1.25$$
Calculate $A_{min}$ numerically and compare with 715 to decide which container requires less material.
Final answers:
- $d = 3l$
- $l = 5$ cm, $d = 15$ cm
- Cuboid surface area = 650 cm²
- Cylinder height $h = \frac{375}{\pi r^2}$
- Surface area $A = 2\pi r^2 + \frac{750}{r}$
- $\frac{dA}{dr} = 4\pi r - \frac{750}{r^2}$
- Minimizing $A$ gives $r = \sqrt[3]{\frac{375}{2\pi}}$
- Compare total materials to choose container.
Container Optimization 95278B
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