If x+y =16 and $x^{2} + y^{2}$ is minimum, then the values of x and y are

3, 13

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4, 12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6, 10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

8, 8

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 8, 8
$x + y = 16 \Rightarrow y = 16 - x \Rightarrow x^{2} + y^{2} = x^{2} + (16 - x)^{2}$
Let $z = x^{2} + (16 - x)^{2} \Rightarrow z^{'} = 4 x - 32$
To be minimum of z, z' = 0, .
Therefore $4 x - 32 = 0 \Rightarrow x = 8 \Rightarrow y = 8$

Suggest Corrections

Similar questions

x + y = 12

x^{2} + y^{2}

x^{2} + 2 x y - y^{2} = 6,

{(x^{2} + y^{2})}^{2} .

x y (x^{2} - y^{2}) = x^{2} + y^{2}

x^{2} + y^{2}

f (x, y) = x y^{2}

x

y

x^{2} + y^{2} = 9

f (x, y)

If x and y are the maximum and minimum values of sin $θ$ + cos $θ$ , find the value of $x^{2}$ + $y^{2}$

Join BYJU'S Learning Program