You can put this solution on YOUR website! 2 positive numbers the sum of whose squares is 16 and whose product is as large as possible
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Equations:
x^2 + y^2 = 16
Product = xy
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y = sqrt(16-x^2)
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Product = x*sqrt(16-x^2)
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Take the derivative
dP/dx = x[(1/2)(16-x^2)^(-1/2)(-2x)] + [sqrt(16-x^2)]
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dP/dx = [-x/sqrt(16-x^2)] + [sqrt(16-x^2)]
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dP/dx = [-x + 16-x^2]/sqrt(16-x^2)
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Solve -x+16-x^2 = 0
x^2+x-16 = 0
x = [-1+-sqrt(1+4*16)]/2
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x = [-1+-sqrt(65)]/2
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Positive solution:
x = 3.53
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Substitute into x^2 + y^2 = 16
y = sqrt(16-x^2)
y = sqrt(16-12.4689)
y = 1.8791
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Cheers,
Stan H.
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