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IB MAI HL Calculus Topic 5.1 Differentiation (id: a4348ab4a)

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admin 发表于 2024-2-27 21:27:14 | 显示全部楼层 |阅读模式
本题目来源于试卷: IB MAI HL Calculus Topic 5.1 Differentiation,类别为 IB数学

[填空题]
A steel pail is made in tm,zseuw4b599:hj x 9xpy .bvihe shape of a cylinder wi8ggr x((*szjns4r0 l cth an internal height h \mathrl sgs 4rzx( 0(n8*cjgrm{~cm} and internal base radius r \mathrm{~cm} .

The steel pail has an open top. The inner surfaces of the pail are to be coated with a protective resin.
1. Write down a formula for A , the internal surface area to be coated.

The volume of the steel pail is $10000 \mathrm{~cm}^{3}$ .
A=a$\pi rh$ + $\pi r^b$ ; a=  ,b=  .
2. Write down, in terms of r and h , an equation for the volume of this steel pail.
3. Show that $A=\pi r^{2}+\frac{20000}{r}$ .

The steel pail is designed so that the area to be coated is minimised.
4. Find $\frac{\mathrm{d} A}{\mathrm{~d} r}$ =-$\frac{a}{r^2}$+b$\pi r$; a=  ,b=  .
5. Using your answer to part (d), find the value of r which minimizes A .
r≈   cm
6. Hence, find the value of this minimum area, correct to the nearest $\mathrm{cm}^{2}$ .
A≈  cm$^2$
One can of protective resin coats a surface area of $350 \mathrm{~cm}^{2}$ .

7. Find the minimum number of cans of protective resin required to coat the area found in part (f).
n≈   cans




参考答案:
空格1: 2空格2: 2空格3: 20000空格4: 2空格5: 14.7±0.1空格6: 2039±1空格7: 6±0.3


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