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

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Aicharya · Accepted Answer

Let the two numbers be a and b . $\begin{array}{ll} \therefore & a+b=3 \\ \therefore & b=3-a...(i) \end{array}$ $\therefore \quad$ Product of first number and square of second number $(p)=a b^2$ $\mathrm{p}=\mathrm{a}(3-\mathrm{a})^2$ $\ldots[$ From (i)] $\begin{array}{ll} \therefore & p=9 a-6 a^2+a^3 ...(ii)\\ \therefore & \frac{d p}{d a}=9-12 a+3 a^2 \\ \therefore & \frac{d^2 p}{d a^2}=-12+6 a \end{array}$
Now, $\frac{\mathrm{dp}}{\mathrm{da}}=0 \Rightarrow 3 \mathrm{a}^2-12 \mathrm{a}+9=0$ $\begin{aligned} & \Rightarrow(3 a-9)(a-1)=0 \\ & \Rightarrow a=3 \text { or } a=1 \end{aligned}$ $\left.\frac{\mathrm{d}^2 p}{d \mathrm{da}^2}\right|_{a=3}=6\gt0 \text { and }\left.\frac{\mathrm{d}^2 p}{\mathrm{da}^2}\right|_{\mathrm{a}=1}=-6 \lt 0$ $\therefore \quad$ Maximum value of p is at $\mathrm{a}=1$ $\therefore \quad$ Maximum value of the product $=4$ ...[From (ii)]

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

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