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Divide the water and theDivide the number 84 Onto duro parte such that the produsof one Poud and the square of the Other is maximum.Let the two paudas be o anel 84-2 |
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Answer» Let the two parts be x and y.x+y = 84⇒ y = (84-x) You need to maximise the product of "one part" and "the square of other"Let one part = (84-x)square ofother part = (x)² = x²product, p = x²×(84-x) = 84x² - x³ dp/dx = 0d/dx(84x² - x³) = 084*2x - 3x² = 03x(56 - x) = 0x = 0, 56 For maximum,d^2 p/dx^2 < 0d^2 (84x² - x³)/dx = d/dx(168x - 3x²)= 168 - 6x at x = 0168 - 6x = 168 - 0 = 168 > 0 at x = 56168 - 6x = 168 - 6*56 = - 168 <0 So x=56,y = 84 - x = 84 - 56 = 28 So the two parts are (28,56) |
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