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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(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:

Similar Questions

The sum of the lengths of the hypotenuse and one side of a right-angled triangle is constant. The area of the triangle will be maximum if the angle between them is:

The maximum value of $\frac{\log x}{x}, 0 < x < \infty$ is

$A$ box is to be made with a square base and an open top. If the area of the material used is $48 \, m^2$,then the maximum volume of the box is ........... $m^3$.

If $y=a \log x+b x^2+x$ has its extreme value at $x=-1$ and $x=2$,then the value of $a+b$ is

Let $P(x)$ be a polynomial of degree $3$ having an extreme value at $x=1$. If $\lim _{x \rightarrow 0}\left(\frac{P(x)+4}{x^2}+2\right)=6$,then $\left(\frac{d P}{d x}\right)_{x=\frac{1}{2}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module