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4x+3y=10, then minimum value of x2+y2 is

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Given, 4x+3y =10

We have to find the minimum value of S = x^{2}+y^{2}

x = \frac{10-3y}{4}

putting value of x in the S,

S =\left ( \frac{10-3y}{4} \right )^{2}+y^{2}

differentiate wrt y

\frac{\mathrm{d} S}{\mathrm{d} y} =-\frac{2\times 3}{4}\left ( \frac{10-3y}{4} \right )+2y

This will be zero for maxima/minima,

\Rightarrow -\frac{2\times 3}{4}\left ( \frac{10-3y}{4} \right )+2y =0

\Rightarrow y =\frac{6}{5}

Find the value of x  by, x = \frac{10-3\times \frac{6}{5}}{4} = \frac{8}{5}

Now minimum value of S, S_{min} =\left ( \frac{6}{5} \right )^{2} + \left ( \frac{8}{5} \right )^{2} = \frac{100}{25} = 4

 

 

Posted by

Abhishek Sahu

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