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Algebra Level 5

P = ( a b ) 2 + ( 8 b 2 25 a ) 2 P=(a-b)^2 + \left(\sqrt {8-b^2} -\frac {25}a \right)^2

P P is defined above for some real numbers a a and b b . Find the minimum value of P P .


The answer is 18.

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2 solutions

I have a geometric solution.P represents the square of the distance between any 2 points on the curves xy=25 and the portion of the circle x 2 x^2 + + y 2 y^2 = = 8 8 above the X-axis respectively.Hence,we want the square of shortest distance between the two curves.Symmetrically, the shortest distance is along y=x.The square of the shortest distance can then be computed,which comes out to be 18.

Same approach!

But after finding a=5 and b = 2, I found the distance between the two points and got ( 18 ) \sqrt(18) and my immediate reaction was reveal solution and then I realized that we had to find square of the distance(--facepalm--)

Harsh Shrivastava - 4 years, 11 months ago

It happens with everyone.

Indraneel Mukhopadhyaya - 4 years, 10 months ago

By AM - GM inequality ( a b ) 2 + ( 8 b 2 25 a ) 2 2 ( ( a b ) ( 8 b 2 25 a ) ) 2 \frac{(a - b)^2 + (\sqrt{8 - b^2} - \frac{25}{a})^{2}}{2} \ge \sqrt{((a - b)(\sqrt{8 - b^2} - \frac{25}{a}))^{2}} and the inequality is only one equality when ( a b ) 2 = ( 8 b 2 25 a ) 2 a b = ± ( 8 b 2 25 a ) (a - b)^2 = (\sqrt{8 - b^2} - \frac{25}{a})^2 \Rightarrow a - b = \pm (\sqrt{8 - b^2} - \frac{25}{a}) \Rightarrow we have two equations, let's choose one: a b = 8 b 2 + 25 a a - b = - \sqrt{8 - b^2} + \frac{25}{a} .This implies that ( a , b ) = ( 5 , 2 ) (a,b) = (5,2) is a solution and then we can get a minimum for P = 18 P = 18 ....

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