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JEE Main202117 Mar 2021Morning ShiftMathematicsApplication of DerivativesActual

The maximum value of z in the following equation z = 6 x y + y 2 , where 3 x + 4 y ≤ 100 and 4 x + 3 y ≤ 75 for x ≥ 0 and y ≥ 0 is

Correct answer

904

Step-by-step solution

z = 6 x y + y 2 = y ( 6 x + y ) 3 x + 4 y ≤ 100   … i 4 x + 3 y ≤ 75   … ii x ≥ 0 y ≥ 0 x ≤ 75 - 3 y 4 Given z = y ( 6 x + y ) ⇒ z ≤ y 6 · 75 - 3 y 4 + y ⇒ z ≤ 1 2 225 y - 7 y 2 Now range of 1 2 225 y - 7 y 2 is ( - ∞ , ( 225 ) 2 2 × 4 × 7 ] So, z ≤ 1 2 225 y - 7 y 2 ≤ ( 225 ) 2 2 × 4 × 7 = 50625 56 ≈ 904 . 0178 ≈ 904 . 02 It will be attained at y = 225 14

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