question_answer

        

For the curve \[y=4{{x}^{3}}-2{{x}^{5}},\] find all the point on the curve at which the tangent passes through the origin,

OR

Show that of all the rectangles with a given perimeter, the square has the largest area.

Answer:

Given curve is \[y=4{{x}^{3}}-2{{x}^{5}}\]                ?(i)

Let any point on the curve be \[({{x}_{1}},\,\,{{y}_{1}}).\]

\[\therefore \]      \[{{y}_{1}}=4x_{1}^{3}-2x_{1}^{5}\]                                   ?(ii)

On differentiating both sides of Eq. (i), we get

\[\frac{dy}{dx}=12{{x}^{2}}-10{{x}^{4}}\]

Equation of tangent at point \[({{x}_{1}},\,\,{{y}_{1}})\] is

\[y-{{y}_{1}}={{\left( \frac{dy}{dx} \right)}_{({{x}_{1}},\,{{y}_{1}})}}(x-{{x}_{1}})\]

\[\therefore \]      \[y-{{y}_{1}}=[12{{({{x}_{1}})}^{2}}-10{{({{x}_{1}})}^{4}}](x-{{x}_{1}})\]

Since, it passes through the origin.

\[\therefore \]      \[0-{{y}_{1}}=(12x_{1}^{2}-10x_{1}^{4})(0-{{x}_{1}})\]

\[\Rightarrow \]   \[{{y}_{1}}=(12x_{1}^{2}-10x_{1}^{4}){{x}_{1}}\]             ?(iii)

From Eqs. (ii) and (iii), we get

x\[(12x_{1}^{2}-10x_{1}^{4}){{x}_{1}}=4x_{1}^{3}-2x_{1}^{5}\]

\[\Rightarrow \]   \[2x_{1}^{3}(6-5x_{1}^{2})=2x_{1}^{3}(2-x_{1}^{2})\]

\[\Rightarrow \] \[2x_{1}^{3}(4-4x_{1}^{2})=0\] \[\Rightarrow \] \[{{x}_{1}}=0\] or \[4-4x_{1}^{2}=0\]

\[\Rightarrow \]   \[{{x}_{1}}=0\] or \[{{x}_{1}}=\pm 1\]

Now,     \[{{x}_{1}}=0\] \[\Rightarrow \] \[{{y}_{1}}=0\]

\[\Rightarrow \] \[{{x}_{1}}=1\] \[\Rightarrow \] \[{{y}_{1}}=4{{(1)}^{3}}-2{{(1)}^{5}}=4-2=2\]

and       \[{{x}_{1}}=-\,1\] \[\Rightarrow \] \[{{y}_{1}}=4{{(-1)}^{3}}-2{{(-1)}^{5}}\]

\[=-\,4+2=-\,2\]

Hence, all points on the curve at which tangent passes through origin are (0, 0), (1, 2) and \[(-\,1,\,\,-2).\]

OR

Let x and y be the lengths of two sides of a rectangle. Also, let P denotes the perimeter and A denotes the area of rectangle.

Given that,        \[P=2(x+y)\]

[\[\therefore \] perimeter of rectangle \[=2(l+b)\]]

\[\Rightarrow \]   \[P=2x+2y\] \[\Rightarrow \] \[y=\frac{P-2x}{2}\]    ?(i)

Also, we know that area of rectangle is given by

A = xy

\[\Rightarrow \]   \[A=x\left( \frac{P-2x}{2} \right)\]                           [from Eq. (i)]

\[\Rightarrow \]   \[A=\frac{Px-2{{x}^{2}}}{2}\]

On differentiating sides w.r.t. x, we get

\[\frac{dA}{dx}=\frac{P-4x}{2}\]

Now, for maxima and minima, put \[\frac{dA}{dx}=0\]

\[\therefore \]      \[\frac{P-4x}{2}=0\] \[\Rightarrow \] P = 4x

\[\Rightarrow \]   \[2x+2y=4x\]             \[[\because \,\,\,P=2x+2y]\]

\[\Rightarrow \]   x = y

\[\because \] x = y, so the rectangle is a square.

Also, \[\frac{{{d}^{2}}A}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{P-4x}{2} \right)=-\frac{4}{2}=-\,2<0\]

\[\therefore \] A is maximum

Hence, area is maximum when rectangle is a square.