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

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(B) Let the two numbers be $a$ and $b$. Given that $a + b = 3$,so $b = 3 - a$. Let the product be $P = a \cdot b^2$. Substituting the value of $b$,we

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$4$

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(B) Let the two numbers be $a$ and $b$. Given that $a + b = 3$,so $b = 3 - a$. Let the product be $P = a \cdot b^2$. Substituting the value of $b$,we get $P = a(3 - a)^2 = a(9 - 6a + a^2) = a^3 - 6a^2 + 9a$. To find the maximum value,we differentiate $P$ with respect to $a$: $\frac{dP}{da} = 3a^2 - 12a + 9$. Setting $\frac{dP}{da} = 0$,we get $3(a^2 - 4a + 3) = 0$,which implies $3(a - 1)(a - 3) = 0$. So,$a = 1$ or $a = 3$. Now,find the second derivative: $\frac{d^2P}{da^2} = 6a - 12$. For $a = 1$,$\frac{d^2P}{da^2} = 6(1) - 12 = -6 For $a = 3$,$\frac{d^2P}{da^2} = 6(3) - 12 = 6 > 0$,so $P$ is minimum at $a = 3$. Substituting $a = 1$ into the product formula: $P = 1(3 - 1)^2 = 1(2)^2 = 4$. Thus,the maximum value is $4$.

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

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