Question
How should we choose two numbers, each greater than or equal to −2, whose sum so that the sum of the first and the cube of the second is minimum?
✓
Answer
Let the two number be x and y. Then,
$\text{x},\text{y}>-2 \ \text{and}\ \text{x}+\text{y}=\frac{1}{2}....(\text{i})$
Now, $\text{z}=\text{x}+\text{y}^{3}$
$\Rightarrow \text{z}=\text{x}+\Big(\frac{1}{2}-\text{x}\Big)^{3}$ [From eq. (i)]
$\Rightarrow\frac{\text{dz}}{\text{dx}}=1+3\Big(\frac{1}{2}-\text{x}\Big)^{2}$
For maximum or minimum values of z, We must have $\Rightarrow\frac{\text{dz}}{\text{dx}}=0$
$\Rightarrow1+3\Big(\frac{1}{2}-\text{x}\Big)^{2}=0$
$\Rightarrow\Big(\frac{1}{2}-\text{x}\Big)^{2}=\frac{1}{3}$
$\Rightarrow\Big(\frac{1}{2}-\text{x}\Big)=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{1}{2}\pm\frac{1}{\sqrt{3}}$
$\frac{\text{d}^{2}\text{z}}{d\text{x}^{2}}=-36\text{x}$
At $\text{x}=\frac{1}{2}\pm\frac{1}{\sqrt{3}}$.
$\text{x},\text{y}>-2 \ \text{and}\ \text{x}+\text{y}=\frac{1}{2}....(\text{i})$
Now, $\text{z}=\text{x}+\text{y}^{3}$
$\Rightarrow \text{z}=\text{x}+\Big(\frac{1}{2}-\text{x}\Big)^{3}$ [From eq. (i)]
$\Rightarrow\frac{\text{dz}}{\text{dx}}=1+3\Big(\frac{1}{2}-\text{x}\Big)^{2}$
For maximum or minimum values of z, We must have $\Rightarrow\frac{\text{dz}}{\text{dx}}=0$
$\Rightarrow1+3\Big(\frac{1}{2}-\text{x}\Big)^{2}=0$
$\Rightarrow\Big(\frac{1}{2}-\text{x}\Big)^{2}=\frac{1}{3}$
$\Rightarrow\Big(\frac{1}{2}-\text{x}\Big)=\pm\frac{1}{\sqrt{3}}$
$\Rightarrow\text{x}=\frac{1}{2}\pm\frac{1}{\sqrt{3}}$
$\frac{\text{d}^{2}\text{z}}{d\text{x}^{2}}=-36\text{x}$
At $\text{x}=\frac{1}{2}\pm\frac{1}{\sqrt{3}}$.
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