If the sum of two numbers is 3,then the maxim | vedclass

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(A) Let the first number be $3 - x$ and the second number be $x$.We want to maximize the product $P(x) = (3 - x)x^2 = 3x^2 - x^3$.To find the critical

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$4$

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(A) Let the first number be $3 - x$ and the second number be $x$.We want to maximize the product $P(x) = (3 - x)x^2 = 3x^2 - x^3$.To find the critical points,we differentiate $P(x)$ with respect to $x$:$P'(x) = 6x - 3x^2$.Setting $P'(x) = 0$,we get $3x(2 - x) = 0$,which gives $x = 0$ or $x = 2$.Now,we find the second derivative $P''(x) = 6 - 6x$.For $x = 2$,$P''(2) = 6 - 6(2) = -6 Since the second derivative is negative at $x = 2$,the function has a local maximum at $x = 2$.The maximum value is $P(2) = (3 - 2)(2^2) = 1 	imes 4 = 4$.

If the sum of two numbers is $3$,then the maximum value of the product of the first and the square of the second is:

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