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.
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