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in Mathematics by (31.4k points)
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The maximum value of z in the following equation z = 6xy + y2, where 3x + 4y \(\leq\) 100 and 4x + 3y \(\leq\) 75 for x \(\geq\) 0 and y \(\geq\) 0 is _______ .

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Answer is (904)

by (10 points)
I can't understand  Why z ≤ 1/2(225y-y²) ≤ (225)²/2.4.7
(225)²/2.4.7 kaahan se aaya???
If you know Please reply
by (48.4k points)
z ≤ 1/2 (225 - 7y^2)
Let f(y) = 225y - 7y^2
f'(y) = 225 - 14y
f'(y) = 0 gives  y = 225/14
f''(y) = -14 < 0
∴ y = 225/14 is point of maxima of f(y).
∴ f(225/14) is maximum value
Maximum value = 225 x 225/14 - 7(225/14)(225/14)
= (225)^2/14 (1 - 1/2)
= (225)^2/28
= (225)^2/4 x 7
Hence, 225y - 7y^2 ≤ (225)^2/4 x 7
∴ z ≤ 1/2 (225 - 7y^2) ≤ (225)^2/ 2 x 4 x 7

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