Maxima and Minima — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima5 Marks
Question
Find the largest possible area of a right angled triangle whose hypotenuse is 5cm long.
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Answer
Let the base of right angled triangle be x and its height be y, Then,
$\text{x}^{2}+\text{y}^{2}=5^{2}$
$\Rightarrow\text{y}^{2}=25-\text{x}^{2}$
$\Rightarrow\text{y}=\sqrt{25-\text{x}^{2}}$
As, thr area of the triangle, $\text{A}=\frac{1}{2}\times\text{x}\times\text{y}$
$\Rightarrow\text{A}(\text{x})=\frac{1}{2}\times\text{x}\times\sqrt{25-\text{x}^{2}}$
$\Rightarrow\text{A}(\text{x})=\frac{\text{x}\sqrt{25-\text{x}^{2}}}{2}$
$\Rightarrow\text{A}'(\text{x})=\frac{\sqrt{25-\text{x}^{2}}}{2}+\frac{\text{x}(-2\text{x})}{4\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}'(\text{x})=\frac{\sqrt{25-\text{x}^{2}}}{2}+\frac{\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}'(\text{x})=\frac{25-\text{x}^{2}-\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}'(\text{x})=\frac{25-2\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}$
For maxima or minima, We must have f'(x) = 0
A'(x) = 0
$\Rightarrow\frac{25-2\text{x}^{2}}{2\sqrt{25-\text{x}^{2}}}=0$
$\Rightarrow 25-2\text{x}^{2}=0$
$\Rightarrow 2\text{x}^{2}=25$
$\Rightarrow \text{x}=\frac{5}{\sqrt{2}}$
So, $\text{y}=\sqrt{25-\frac{25}{2}}$
$=\sqrt{\frac{50-25}{2}}$
$=\sqrt{\frac{25}{2}}$
$={\frac{5}{\sqrt{2}}}$
Also, $\text{A}'(\text{x})=\frac{\Bigg[-4\text{x}\sqrt{25-\text{x}^{2}}-\frac{(25-2\text{x}^{2})(-2\text{x})}{2\sqrt{25-\text{x}^{2}}}\Bigg]}{25-\text{x}^{2}}$
$=\frac{\Bigg[\frac{-4\text{x}(25-\text{x}^{2})+(25\text{x}-2\text{x}^{3})}{\sqrt{25-\text{x}^{2}}}\Bigg]}{25-\text{x}^{2}}$
$=\frac{-100\text{x}+4\text{x}^{3}+25\text{x}-2\text{x}^{3}}{(25-\text{x}^{2})\sqrt{25-\text{x}^{2}}}$
$=\frac{-75\text{x}+2\text{x}^{3}}{(25-\text{x}^{3})\sqrt{25-\text{x}^{2}}}$
$\Rightarrow\text{A}''\Big(\frac{5}{\sqrt{2}}\Big)=\frac{-75\Big(\frac{5}{\sqrt{2}}\Big)+2\Big(\frac{5}{\sqrt{2}}\Big)^{3}}{\Bigg(25-\Big(\frac{5}{\sqrt{2}}\Big)^{2}\Bigg)^\frac{3}{2}}<0$
So, $\text{x}=\Big(\frac{5}{\sqrt{2}}\Big)$ is point of maxima.
$\therefore$ The largest possible area of triangle $=\frac{1}{2}\times\Big(\frac{5}{\sqrt{2}}\Big)\times\Big(\frac{5}{\sqrt{2}}\Big)=\frac{25}{4}$ square units.
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