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JEE Main 2021 · MathsNumerical answerNumerical valueMediumMulti-step

JEE Main 17 March 2021, Shift 1, Maths Q81

Question 81 of 90 in this shift, Maths question 21 of 30, Section B.

The maximum value of

z

in the following equation

z=6xy+y^{2}

, where

3x+4y\leq100

and

4x+3y\leq75

for

x\geq0

and

y\geq0

is __________.

Official answer

904

NTA final key.

Chapter: Application of Derivatives
Topic: No close match in our taxonomy
Idea tested: No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.