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 | algebra |
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x + y = 8 and P = 5x^2 + 11y^2, where x, y > 0. What is the minimum possible value of P? ans.220(pls explain) |
| Re: algebra |
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Hi Rahul, Yes ans is 220 as follows. P=5x^2 + 11y^2 put y=8-x we will get P=16x^2-176x+704 So P will have a minimum value.And that min value will be at x=(-b)/2a = 176/32 = 11/2 Putting x=11/2 we get y=5/2 put these values in P=5x^2 + 11y^2 We will get Pmin=220. |
| Re: algebra |
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rahul, alternate method by differentitaion could be, p= 5x^2+11(8-x)^2 further dp/dx = 32x-176 =>dp/dx=0gives x=11/2 d2p/dx2 =32 which is positive, means minima exists at dp/dx, put x=11/2 and get the value as 220. you can also follow method od quadratic equations as mentioned by Ramesh, cheers ramesh!! |
