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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? - CBSE (Arts) Class 12 - Mathematics

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?

Solution

$\text { Let the two numbers bexandy. Then },$

$x, y > - 2 \text { and }x + y = \frac{1}{2} ........ (1)$

$\text { Now,}$

$z = x + y^3$

$\Rightarrow z = x + \left( \frac{1}{2} - x \right)^3 .........\left[ \text { From eq } . \left( 1 \right) \right]$

$\Rightarrow \frac{dz}{dx} = 1 + 3 \left( \frac{1}{2} - x \right)^2$

$\text { For maximum or minimum values of z, we must have }$

$\frac{dz}{dx} = 0$

$\Rightarrow 1 + 3 \left( \frac{1}{2} - x \right)^2 = 0$

$\Rightarrow \left( \frac{1}{2} - x \right)^2 = \frac{1}{3}$

$\Rightarrow \left( \frac{1}{2} - x \right) = \pm \frac{1}{\sqrt{3}}$

$\Rightarrow x = \frac{1}{2} \pm \frac{1}{\sqrt{3}}$

$\frac{d^2 z}{d x^2} = 6\left( \frac{1}{2} - x \right)$

$\Rightarrow \frac{d^2 z}{d x^2} = 3 - 6x$

$\text { At } x = \frac{1}{2} \pm \frac{1}{\sqrt{3}}:$

$\frac{d^2 z}{d x^2} = 3 - 6\left( \frac{1}{2} + \frac{1}{\sqrt{3}} \right)$

$\Rightarrow \frac{- 6}{\sqrt{3}} < 0$

$\text { Thus, z is maximum when x } = \frac{1}{2} + \frac{1}{\sqrt{3}} .$

$\text { At x } = \frac{1}{2} - \frac{1}{\sqrt{3}}:$

$\frac{d^2 z}{d x^2} = 3 - 6\left( \frac{1}{2} - \frac{1}{\sqrt{3}} \right)$

$\Rightarrow \frac{6}{\sqrt{3}} > 0$

$\text { Thus, z is minimum when x }= \frac{1}{2} - \frac{1}{\sqrt{3}} .$

$x + y = \frac{1}{2}$

$\text { Substituting the value of x in eq }.\left( 1 \right),\text { we get }$

$y = - \frac{1}{2} + \frac{1}{\sqrt{3}} + \frac{1}{2}$

$y = \frac{1}{\sqrt{3}}$

$\text { So, the required two numbers are } \left( \frac{1}{2} - \frac{1}{\sqrt{3}} \right) \text { and } \frac{1}{\sqrt{3}} .$

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Solution 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? Concept: Graph of Maxima and Minima.
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