Question:
The sum of two numbers is $10$. Their product will be maximum when they are
The sum of two numbers is $10$. Their product will be maximum when they are
Updated On: Jun 14, 2022
- $3, 7$
- $4, 6$
- $5, 5$
- $8, 2$
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The Correct Option is C
Solution and Explanation
Let one number be $x$ and second number be $(10 - x)$.
According to the question,
$P = x(10 - x)$, where $P$ is the product of numbers on differentiating both sides w.r.t. ?$x$' we get
$\frac{dP}{dx} = x \frac{d}{dx} (10 - x) + ( 10 - x) \times \frac{d}{dx} (x) $
$ = x \times (-1) + (10 - x) \times 1$
$ = - x - x + 10$
$\frac{dP}{dx} = - 2x + 10 \,\,\,...(i)$
For maximum or minimum value,
$\frac{dP}{dx} = 0$
$\Rightarrow -2x + 10 = 0$
$\Rightarrow x = 5$
Now, on differentiating E $(i)$ w.r.t. $'x'$ we get
$\frac{d^2P}{dx^2} = -2 < 0$
$\therefore P$ is maximum at $ x = 5$
Hence, numbers are $5, 5$.
According to the question,
$P = x(10 - x)$, where $P$ is the product of numbers on differentiating both sides w.r.t. ?$x$' we get
$\frac{dP}{dx} = x \frac{d}{dx} (10 - x) + ( 10 - x) \times \frac{d}{dx} (x) $
$ = x \times (-1) + (10 - x) \times 1$
$ = - x - x + 10$
$\frac{dP}{dx} = - 2x + 10 \,\,\,...(i)$
For maximum or minimum value,
$\frac{dP}{dx} = 0$
$\Rightarrow -2x + 10 = 0$
$\Rightarrow x = 5$
Now, on differentiating E $(i)$ w.r.t. $'x'$ we get
$\frac{d^2P}{dx^2} = -2 < 0$
$\therefore P$ is maximum at $ x = 5$
Hence, numbers are $5, 5$.
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives