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

Hint: 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