Advertisements
Advertisements
Question
Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.
Solution
\[\text { Let the base of the right angled triangle be x and its height be y . Then,} \]
\[ x^2 + y^2 = 5^2 \]
\[ \Rightarrow y^2 = 25 - x^2 \]
\[ \Rightarrow y = \sqrt{25 - x^2}\]
\[\text { As, the area of the triangle, }A = \frac{1}{2} \times x \times y\]
\[ \Rightarrow A\left( x \right) = \frac{1}{2} \times x \times \sqrt{25 - x^2}\]
\[ \Rightarrow A\left( x \right) = \frac{x\sqrt{25 - x^2}}{2}\]
\[ \Rightarrow A'\left( x \right) = \frac{\sqrt{25 - x^2}}{2} + \frac{x\left( - 2x \right)}{4\sqrt{25 - x^2}}\]
\[ \Rightarrow A'\left( x \right) = \frac{\sqrt{25 - x^2}}{2} - \frac{x^2}{2\sqrt{25 - x^2}}\]
\[ \Rightarrow A'\left( x \right) = \frac{25 - x^2 - x^2}{2\sqrt{25 - x^2}}\]
\[ \Rightarrow A'\left( x \right) = \frac{25 - 2 x^2}{2\sqrt{25 - x^2}}\]
\[\text { For maxima or minima, we must have } f'\left( x \right) = 0\]
\[A'\left( x \right) = 0\]
\[ \Rightarrow \frac{25 - 2 x^2}{2\sqrt{25 - x^2}} = 0\]
\[ \Rightarrow 25 - 2 x^2 = 0\]
\[ \Rightarrow 2 x^2 = 25\]
\[ \Rightarrow x = \frac{5}{\sqrt{2}}\]
\[\text { So, y } = \sqrt{25 - \frac{25}{2}}\]
\[ = \sqrt{\frac{50 - 25}{2}}\]
\[ = \sqrt{\frac{25}{2}}\]
\[ = \frac{5}{\sqrt{2}}\]
\[\text { Also,} A''\left( x \right) = \frac{\left[ - 4x\sqrt{25 - x^2} - \frac{\left( 25 - 2 x^2 \right)\left( - 2x \right)}{2\sqrt{25 - x^2}} \right]}{25 - x^2}\]
\[ = \frac{\left[ \frac{- 4x\left( 25 - x^2 \right) + \left( 25x - 2 x^3 \right)}{\sqrt{25 - x^2}} \right]}{25 - x^2}\]
\[ = \frac{- 100x + 4 x^3 + 25x - 2 x^3}{\left( 25 - x^2 \right)\sqrt{25 - x^2}}\]
\[ = \frac{- 75x + 2 x^3}{\left( 25 - x^2 \right)\sqrt{25 - x^2}}\]
\[ \Rightarrow A''\left( \frac{5}{\sqrt{2}} \right) = \frac{- 75\left( \frac{5}{\sqrt{2}} \right) + 2 \left( \frac{5}{\sqrt{2}} \right)^3}{\left( 25 - \left( \frac{5}{\sqrt{2}} \right)^2 \right)^\frac{3}{2}} < 0\]
\[So, x = \left( \frac{5}{\sqrt{2}} \right) \text { is point of maxima }. \]
\[ \therefore \text { The largest possible area of the triangle } = \frac{1}{2} \times \left( \frac{5}{\sqrt{2}} \right) \times \left( \frac{5}{\sqrt{2}} \right) = \frac{25}{4} \text { square units }\]
APPEARS IN
RELATED QUESTIONS
f(x)=sin 2x+5 on R .
f(x)=2x3 +5 on R .
f(x) = cos x, 0 < x < \[\pi\] .
`f(x)=sin2x-x, -pi/2<=x<=pi/2`
`f(x) = x/2+2/x, x>0 `.
f(x) = \[x^3 - 2a x^2 + a^2 x, a > 0, x \in R\] .
f(x) = \[x + \frac{a2}{x}, a > 0,\] , x ≠ 0 .
f(x) = (x \[-\] 1) (x \[-\] 2)2.
`f(x) = 3x^4 - 8x^3 + 12x^2- 48x + 25 " in "[0,3]` .
f(x) = (x \[-\] 2) \[\sqrt{x - 1} \text { in }[1, 9]\] .
Find the maximum value of 2x3\[-\] 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [ \[-\] 3, \[-\] 1].
Determine two positive numbers whose sum is 15 and the sum of whose squares is maximum.
Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?
A beam is supported at the two end and is uniformly loaded. The bending moment M at a distance x from one end is given by \[M = \frac{WL}{2}x - \frac{W}{2} x^2\] .
Find the point at which M is maximum in a given case.
A square piece of tin of side 18 cm is to be made into a box without top by cutting a square from each corner and folding up the flaps to form a box. What should be the side of the square to be cut off so that the volume of the box is maximum? Find this maximum volume.
A tank with rectangular base and rectangular sides, open at the top, is to the constructed so that its depth is 2 m and volume is 8 m3. If building of tank cost 70 per square metre for the base and Rs 45 per square metre for sides, what is the cost of least expensive tank?
Show that the height of the cone of maximum volume that can be inscribed in a sphere of radius 12 cm is 16 cm ?
Show that the maximum volume of the cylinder which can be inscribed in a sphere of radius \[5\sqrt{3 cm} \text { is }500 \pi {cm}^3 .\]
A box of constant volume c is to be twice as long as it is wide. The material on the top and four sides cost three times as much per square metre as that in the bottom. What are the most economic dimensions?
A particle is moving in a straight line such that its distance at any time t is given by S = \[\frac{t^4}{4} - 2 t^3 + 4 t^2 - 7 .\] Find when its velocity is maximum and acceleration minimum.
Write the point where f(x) = x log, x attains minimum value.
Find the least value of f(x) = \[ax + \frac{b}{x}\], where a > 0, b > 0 and x > 0 .
Write the maximum value of f(x) = x1/x.
If \[ax + \frac{b}{x} \frac{>}{} c\] for all positive x where a,b,>0, then _______________ .
The minimum value of f(x) = \[x4 - x2 - 2x + 6\] is _____________ .
The function f(x) = \[\sum^5_{r = 1}\] (x \[-\] r)2 assumes minimum value at x = ______________ .
At x= \[\frac{5\pi}{6}\] f(x) = 2 sin 3x + 3 cos 3x is ______________ .
The least value of the function f(x) = \[x3 - 18x2 + 96x\] in the interval [0,9] is _____________ .
The maximum value of f(x) = \[\frac{x}{4 - x + x^2}\] on [ \[-\] 1, 1] is _______________ .
The least and greatest values of f(x) = x3\[-\] 6x2+9x in [0,6], are ___________ .
If(x) = x+\[\frac{1}{x}\],x > 0, then its greatest value is _______________ .
Let x, y be two variables and x>0, xy=1, then minimum value of x+y is _______________ .
f(x) = 1+2 sin x+3 cos2x, `0<=x<=(2pi)/3` is ________________ .
The maximum value of f(x) = \[\frac{x}{4 + x + x^2}\] on [ \[-\] 1,1] is ___________________ .
