计算满足不等式约束的面积大小

求满足形如 $|ax+by|+|cx+dy|\le k$ 的不等式的面积 ​

解 ​

换元 ​

$$u=ax+by,v=cx+dy$$

换元后面积 ​

易知该面积为 $S_{tran}=\frac{(2k{)}^2}{2}=2k^2$

面积变换 ​

Lemma I ​

$\mathbf{T}$ 为变换矩阵.

$$\begin{array}{rl}\mathbf{T}(u+\mathrm{\Delta }u,v)-\mathbf{T}(u,v)& =\frac{\mathbf{T}(u+\mathrm{\Delta }u,v)-\mathbf{T}(u,v)}{\mathrm{\Delta }u}\mathrm{\Delta }u\\ & \approx \frac{\partial \mathbf{T}}{\partial u}(u,v)\mathrm{\Delta }u\\ \mathbf{T}(u,v+\mathrm{\Delta }v)-\mathbf{T}(u,v)& =\frac{\mathbf{T}(u,v+\mathrm{\Delta }v)-\mathbf{T}(u,v)}{\mathrm{\Delta }v}\mathrm{\Delta }v\\ & \approx \frac{\partial \mathbf{T}}{\partial v}(u,v)\mathrm{\Delta }v\end{array}$$$$\begin{array}{rl}\mathrm{\Delta }S& \approx \Vert \frac{\partial \mathbf{T}}{\partial u}\mathrm{\Delta }u\times \frac{\partial \mathbf{T}}{\partial v}\mathrm{\Delta }v\Vert =\Vert \frac{\partial \mathbf{T}}{\partial u}\times \frac{\partial \mathbf{T}}{\partial v}\Vert \mathrm{\Delta }u\mathrm{\Delta }v\\ & =\left|\det \left(\left[\begin{array}{cc}\frac{\partial T_1}{\partial u}& \frac{\partial T_2}{\partial u}\\ \frac{\partial T_1}{\partial v}& \frac{\partial T_2}{\partial v}\end{array}\right]\right)\right|\mathrm{\Delta }u\mathrm{\Delta }v.\end{array}$$

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根据 Lemma I

$$\left(\begin{array}{c}du\\ dv\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)\left(\begin{array}{c}dx\\ dy\end{array}\right)$$$$\therefore S_{orig}=\frac{S_{tran}}{|\det \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|}=\frac{S_{tran}}{|ad-bc|}=\frac{2k^2}{|ad-bc|}$$

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习题 ​

Easy: $|x+2y|+|3x+4y|\le 5$ HINT:

Hard: $|x+2y^2|+|3x^2+4y|\le 5$ HINT: (Desmos 和 Geogebra 都战败了, 只有 Mathematia 画出了图像