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The two positive numbers whose sum is 16 and the sum of whose squares is minimum then the positive numbers are:

Solution

✅ Correct Option: 3

Let the two positive numbers be xx and yy.

Given constraints:

x+y=16x + y = 16

The sum of squares needs to be minimized: S=x2+y2S = x^2 + y^2


From the constraint x+y=16x + y = 16, express y=16−xy = 16 - x.

Substitute into the sum of squares:

S=x2+(16−x)2S = x^2 + (16-x)^2

S=x2+256−32x+x2S = x^2 + 256 - 32x + x^2

S=2x2−32x+256S = 2x^2 - 32x + 256


To find the minimum, differentiate with respect to xx:

dSdx=4x−32\dfrac{dS}{dx} = 4x - 32

Set the derivative equal to zero:

4x−32=04x - 32 = 0

4x=324x = 32

x=8x = 8


Since x=8x = 8 and x+y=16x + y = 16:

y=16−8y = 16 - 8

y=8y = 8


The sum of squares for various pairs with sum 16:

For (6,10)(6, 10): 62+102=36+100=1366^2 + 10^2 = 36 + 100 = 136

For (7,9)(7, 9): 72+92=49+81=1307^2 + 9^2 = 49 + 81 = 130

For (8,8)(8, 8): 82+82=64+64=1288^2 + 8^2 = 64 + 64 = 128

For (5,11)(5, 11): 52+112=25+121=1465^2 + 11^2 = 25 + 121 = 146

The minimum sum of squares is 128128, achieved when both numbers equal 88.

Therefore, the two positive numbers are 88 and 88.

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