GO Classes Test Series 2023 | Calculus | Test 1 | Question: 16

Answer 1

Suppose the numbers are $x, x, y$. Then $2 x+y=12$ meaning that $y=12-2 x$, and so the product is $p(x)=x^{2}(12-2 x)=12 x^{2}-2 x^{3}$. Since we must have $0<x<6$ we look for critical $\#$s in this range.

Since $p^{\prime}(x)=24 x-6 x^{2}=6 x(4-x)$, the only critical $\#$ of $p$ in that range is $x=4$, which is local max via the sign diagram for $p^{\prime}$. This means the maximum product is $p(4)=64$.

Comments

@Sachin Mittal 1 sir in this qn, i did like taking out random cases and like tht at last got to  ^{the conclusion like 4,4,4 getting these , other cases fails}

^{can we do like this?} 

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If you know that other values can not exist then you can do it.

As you said that “at last got the conclusion like 4,4,4 getting these.. ”. So it matters “how” you got this conclusion. In the process, if you were sure that there can not be another case then you can do it.

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yes sir i got these value at the very first time when i was calculating but still i checked all options that is other value more than this possible or not and after tht got to final conclusion

thanks sir !