Question

Let positive real numbers $x$ and $y$ be such that $3x + 4y = 14$ . The maximum value of ${x^3}{y^4}$ is
(A) $1056$
(B) $128$
(C) $216$
(D) $432$

Answer 1

First find the value of $y$ in terms of $x$ from the given equation. Then put the value of $y$ in ${x^3}{y^4}$ . Then differentiate this term with respect to $x$ using differentiation results. Then equate it to $0$ and find the value of $x$ . Again differentiate the term with respect to $x$ using differentiation results and then put the value of $x$ obtained. If the value obtained is less than $0$ , then ${x^3}{y^4}$ attains maximum at the value of $x$ obtained. Finally, put this value of $x$ in ${x^3}{y^4}$ and find its maximum value.

Formula used: If $f\left( x \right)$ and $g\left( x \right)$ are differentiable functions and c is a constant.
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{df{{(x)}^n}}}{{dx}} = nf{(x)^{n - 1}}\dfrac{d}{{dx}}f(x)$
$\dfrac{{d\left( c \right)}}{{dx}} = 0$
\dfrac{d}{{dx}}\left\\{ {cf\left( x \right)} \right\\} = c \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right)
$\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$
$\dfrac{d}{{dx}}\left[ {f\left( x \right) \pm g\left( x \right)} \right] = \dfrac{d}{{dx}}f\left( x \right) \pm \dfrac{d}{{dx}}g\left( x \right)$

Complete step-by-step solution:
It is given that positive real numbers $x$ and $y$ are such that $3x + 4y = 14$ .
First find the value of $y$ in terms of $x$ from the given equation.
On simplification, we get
$\Rightarrow$ $y = \dfrac{{14 - 3x}}{4}$ .
Let us consider $S$ be the maximum value of ${x^3}{y^4}$ .
So, we can write it as $S = {x^3}{y^4}$ .
On putting $y = \dfrac{{14 - 3x}}{4}$ in $S = {x^3}{y^4}$
$S = {x^3}{\left( {\dfrac{{14 - 3x}}{4}} \right)^4}$
On simplification, we get
$\Rightarrow S = \dfrac{1}{{256}}{x^3}{\left( {14 - 3x} \right)^4} \ldots \ldots \left( 1 \right)$
Differentiating this equation with respect to $x$ , we get
$\dfrac{{d{\text{S}}}}{{dx}} = \dfrac{d}{{dx}}\left[ {\dfrac{1}{{256}}{x^3}{{\left( {14 - 3x} \right)}^4}} \right]$
Use the property \dfrac{d}{{dx}}\left\\{ {c \times f\left( x \right)} \right\\} = c \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) in above equation where $f\left( x \right)$ is a differentiable function and c is a constant.
$\dfrac{{d{\text{S}}}}{{dx}} = \dfrac{1}{{256}}\dfrac{d}{{dx}}\left[ {{x^3}{{\left( {14 - 3x} \right)}^4}} \right]$
Use the property $\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$ in above equation where $f\left( x \right)$ and $g\left( x \right)$ are differentiable functions.
$\dfrac{{d{\text{S}}}}{{dx}} = \dfrac{1}{{256}}\left[ {{x^3}\dfrac{d}{{dx}}{{\left( {14 - 3x} \right)}^4} + {{\left( {14 - 3x} \right)}^4}\dfrac{d}{{dx}}{x^3}} \right]$
Use the property $\dfrac{{df{{(x)}^n}}}{{dx}} = nf{(x)^{n - 1}}\dfrac{d}{{dx}}f(x)$ and $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ in above equation where $f\left( x \right)$ is a differentiable function.
$\dfrac{{d{\text{S}}}}{{dx}} = \dfrac{1}{{256}}\left[ {{x^3} \times 4{{\left( {14 - 3x} \right)}^3}\left( { - 3} \right) + {{\left( {14 - 3x} \right)}^4} \times 3{x^2}} \right]$
On simplification, we get
$\dfrac{{d{\text{S}}}}{{dx}} = \dfrac{3}{{256}}\left[ {{x^2}{{\left( {14 - 3x} \right)}^4} - 4{x^3}{{\left( {14 - 3x} \right)}^3}} \right]\; \ldots \ldots \left( 2 \right)$
Now to find the maximum value of ${x^3}{y^4}$ , put $\dfrac{{d{\text{S}}}}{{dx}} = 0$ and find the value of $x$
$\dfrac{{d{\text{S}}}}{{dx}} = 0$
So we can write it as,
$\Rightarrow \dfrac{3}{{256}}\left[ {{x^2}{{\left( {14 - 3x} \right)}^4} - 4{x^3}{{\left( {14 - 3x} \right)}^3}} \right] = 0$
Multiply both sides of the equation by $\dfrac{{256}}{3}$ , we get
$\Rightarrow {x^2}{\left( {14 - 3x} \right)^4} - 4{x^3}{\left( {14 - 3x} \right)^3} = 0$
Take ${x^2}{\left( {14 - 3x} \right)^3}$ common in above equation, we get
$\Rightarrow {x^2}{\left( {14 - 3x} \right)^3}\left[ {14 - 3x - 4x} \right] = 0$
On adding the bracket term, we get
$\Rightarrow {x^2}{\left( {14 - 3x} \right)^3}\left[ {14 - 7x} \right] = 0$
On simply we get the value,
$\Rightarrow x = 0,\dfrac{{14}}{3},2$
Again differentiating (2) with respect to $x$ , we get
\Rightarrow \dfrac{{{d^2}{\text{S}}}}{{d{x^2}}} = \dfrac{d}{{dx}}\left\\{ {\dfrac{3}{{256}}\left[ {{x^2}{{\left( {14 - 3x} \right)}^4} - 4{x^3}{{\left( {14 - 3x} \right)}^3}} \right]} \right\\}
Use the property \dfrac{d}{{dx}}\left\\{ {c \times f\left( x \right)} \right\\} = c \times \dfrac{d}{{dx}}\left( {f\left( x \right)} \right) in above equation where $f\left( x \right)$ is a differentiable function and c is a constant.
$\Rightarrow \dfrac{{{d^2}{\text{S}}}}{{d{x^2}}} = \dfrac{3}{{256}}\dfrac{d}{{dx}}\left[ {{x^2}{{\left( {14 - 3x} \right)}^4} - 4{x^3}{{\left( {14 - 3x} \right)}^3}} \right]$
Use the property $\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right)$ in above equation where $f\left( x \right)$ and $g\left( x \right)$ are differentiable functions.
$\Rightarrow \dfrac{{{d^2}{\text{S}}}}{{d{x^2}}} = \dfrac{3}{{256}}\left[ {{x^2}\dfrac{d}{{dx}}{{\left( {14 - 3x} \right)}^4} + {{\left( {14 - 3x} \right)}^4}\dfrac{{d{x^2}}}{{dx}} - 4{x^3}\dfrac{d}{{dx}}{{\left( {14 - 3x} \right)}^3} - 4{{\left( {14 - 3x} \right)}^3}\dfrac{{d{x^3}}}{{dx}}} \right]$
Use the property $\dfrac{{df{{(x)}^n}}}{{dx}} = nf{(x)^{n - 1}}\dfrac{d}{{dx}}f(x)$ and $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ in above equation where $f\left( x \right)$ is a differentiable function.
$\Rightarrow \dfrac{{{d^2}{\text{S}}}}{{d{x^2}}} = \dfrac{3}{{256}}\left[ {{x^2} \times 4{{\left( {14 - 3x} \right)}^3}\left( { - 3} \right) + {{\left( {14 - 3x} \right)}^4}\left( {2x} \right) - 4{x^3} \times 2{{\left( {14 - 3x} \right)}^2}\left( { - 3} \right) - 4{{\left( {14 - 3x} \right)}^3}\left( {3{x^2}} \right)} \right]$
On multiply the term and we get
$\Rightarrow \dfrac{{{d^2}{\text{S}}}}{{d{x^2}}} = \dfrac{3}{{256}}\left[ { - 12{x^2}{{\left( {14 - 3x} \right)}^3} + 2x{{\left( {14 - 3x} \right)}^4} + 24{x^3}{{\left( {14 - 3x} \right)}^2} - 12{x^2}{{\left( {14 - 3x} \right)}^3}} \right]$
On adding the same exponent term and we get,
$\Rightarrow \dfrac{{{d^2}{\text{S}}}}{{d{x^2}}} = \dfrac{3}{{256}}\left[ {2x{{\left( {14 - 3x} \right)}^4} - 24{x^2}{{\left( {14 - 3x} \right)}^3} + 24{x^3}{{\left( {14 - 3x} \right)}^2}} \right]$
Put $x = 2$ in the above equation and check whether the value is less than $0$ or not.
$\Rightarrow {\left( {\dfrac{{{d^2}{\text{S}}}}{{d{x^2}}}} \right)_{x = 2}} = \dfrac{3}{{256}}\left[ {2\left( 2 \right){{\left( {14 - 3\left( 2 \right)} \right)}^4} - 24{{\left( 2 \right)}^2}{{\left( {14 - 3\left( 2 \right)} \right)}^3} + 24{{\left( 2 \right)}^3}{{\left( {14 - 3\left( 2 \right)} \right)}^2}} \right]$
On multiply the term and we get
$\Rightarrow {\left( {\dfrac{{{d^2}{\text{S}}}}{{d{x^2}}}} \right)_{x = 2}} = \dfrac{3}{{256}}\left[ {4 \times {8^4} - 24 \times 4 \times {8^3} + 24 \times 8 \times {8^2}} \right]$
Taking ${8^3}$ as common and we get,
$\Rightarrow \dfrac{{3 \times {8^3}}}{{256}}\left[ {4 \times 8 - 24 \times 4 + 24} \right]$
On multiply the term and square the numerator term we get,
$\Rightarrow 3 \times \dfrac{{512}}{{256}}\left[ {32 - 96 + 24} \right]$
On dividing and add the term we get
$\Rightarrow 3 \times 2\left[ { - 40} \right]$
On multiply and subtract the term and we get
$\Rightarrow 6 \times \left( { - 40} \right)$
Let us multiply the term and we get
$\Rightarrow - 240$
As, ${\left( {\dfrac{{{d^2}{\text{S}}}}{{d{x^2}}}} \right)_{x = 2}} < 0$, so $x = 2$ is a point of local maxima.
i.e., $S$attains maximum value at $x = 2$ .
To find the maximum value of$S$, we have to find the value of $S$ at $x = 2$ .
Put $x = 2$ in equation $\left( 1 \right)$ and find the maximum value of$S$.
${\text{max S}} = \dfrac{1}{{256}}{\left( 2 \right)^3}{\left( {14 - 3\left( 2 \right)} \right)^4}$
On simplify the term and we get,
$= \dfrac{8}{{256}} \times {8^4}$
Let us multiply the term and we get, $f'(x)$
$= 128$

Therefore, The maximum value of ${x^3}{y^4}$ is $\left( B \right)128$ .

Note: Algorithm for determining extreme values of a function:
Steps for determining maxima and minima of $f(x)$
Step 1: Find
Step 2: Find $f''(x)$ . Consider $x = {c_1}$ .
If $f''({c_1}) < 0$ , then $x = {c_1}$ is called the local maximum.
If $f''({c_1}) > 0$ , then $x = {c_1}$ is called the local minimum.
If $f''({c_1}) = 0$ , we must find $f'''(x)$ and substitute in it ${c_1}$ for$x$
If $f'''(x) \ne 0$ , then $x = {c_1}$ is neither a point of local maximum nor a point of local minimum and is called the point of inflection.
Algorithm for determining a function has maximum value at given point:
Let the function be $f(x)$
Step 1: Find $f'(x)$
Step 2: Put $f'(x) = 0$ at given point and find $x$
Step 3: Find $f''(x)$ at $x$ and check it is less than 0.
If $f''(x) < 0$ then the function has maximum value at given point