Advertisements
Advertisements
प्रश्न
A wire of length 20 m is to be cut into two pieces. One of the pieces will be bent into shape of a square and the other into shape of an equilateral triangle. Where the we should be cut so that the sum of the areas of the square and triangle is minimum?
Advertisements
उत्तर
\[\text {Suppose the wire, which is to be made into a square and a triangle, is cut into two pieces of length x and y, respectively . Then}, \]
\[x + y = 20 .......... \left( 1 \right)\]
\[\text { Perimeter of square }, 4\left( Side \right) = x\]
\[ \Rightarrow \text { Side } = \frac{x}{4}\]
\[\text { Area of square = }\left( \frac{x}{4} \right)^2 = \frac{x^2}{16}\]
\[\text { Perimeter of triangle }, 3\left( \text { Side } \right) = y\]
\[ \Rightarrow \text { Side } = \frac{y}{3}\]
\[\text { Area of triangle } = \frac{\sqrt{3}}{4} \times \left( \text { Side } \right)^2 = \frac{\sqrt{3}}{4} \times \left( \frac{y}{3} \right)^2 = \frac{\sqrt{3} y^2}{36}\]
\[\text { Now,} \]
\[z =\text { Area of square + Area of triangle }\]
\[ \Rightarrow z = \frac{x^2}{16} + \frac{\sqrt{3} y^2}{36}\]
\[ \Rightarrow z = \frac{x^2}{16} + \frac{\sqrt{3} \left( 20 - x \right)^2}{36} .................\left[ \text { From eq } . \left( 1 \right) \right]\]
\[ \Rightarrow \frac{dz}{dx} = \frac{2x}{16} - \frac{2\sqrt{3}\left( 20 - x \right)}{36}\]
\[\text { For maximum or minimum values of z, we must have }\]
\[\frac{dz}{dx} = 0\]
\[ \Rightarrow \frac{2x}{16} - \frac{\sqrt{3}\left( 20 - x \right)}{18} = 0\]
\[ \Rightarrow \frac{9x}{4} = \sqrt{3}\left( 20 - x \right)\]
\[ \Rightarrow \frac{9x}{4} + x\sqrt{3} = 20\sqrt{3}\]
\[ \Rightarrow x\left( \frac{9}{4} + \sqrt{3} \right) = 20\sqrt{3}\]
\[ \Rightarrow x = \frac{20\sqrt{3}}{\left( \frac{9}{4} + \sqrt{3} \right)}\]
\[ \Rightarrow x = \frac{80\sqrt{3}}{\left( 9 + 4\sqrt{3} \right)}\]
\[ \Rightarrow y = 20 - \frac{80\sqrt{3}}{9 + 4\sqrt{3}} .............\left[ \text { From eq }. \left( 1 \right) \right]\]
\[ \Rightarrow y = \frac{180}{9 + 4\sqrt{3}}\]
\[ \frac{d^2 z}{d x^2} = \frac{1}{8} + \frac{\sqrt{3}}{18} > 0\]
\[\text { Thus, z is minimum when } x = \frac{80\sqrt{3}}{\left( 9 + 4\sqrt{3} \right)} \text { and }y = \frac{180}{9 + 4\sqrt{3}} . \]
\[\text { Hence, the wire of length 20 cm should be cut into two pieces of lengths }\frac{80\sqrt{3}}{\left( 9 + 4\sqrt{3} \right)} \text { m and } \frac{180}{9 + 4\sqrt{3}} m . \]
Notes
The solution given in the book is incorrrect . The solution here is created according to the question given in the book.
संबंधित प्रश्न
f(x)=2x3 +5 on R .
f(x) = sin 2x, 0 < x < \[\pi\] .
`f(x)=sin2x-x, -pi/2<=x<=pi/2`
f(x) = x4 \[-\] 62x2 + 120x + 9.
`f(x) = 2/x - 2/x^2, x>0`
f(x) = xex.
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) = \[x + \sqrt{1 - x}, x \leq 1\] .
`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 absolute maximum and minimum values of the function of given by \[f(x) = \cos^2 x + \sin x, x \in [0, \pi]\] .
How should we choose two numbers, each greater than or equal to `-2, `whose sum______________ so that the sum of the first and the cube of the second is minimum?
Of all the closed cylindrical cans (right circular), which enclose a given volume of 100 cm3, which has the minimum surface area?
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 rectangle is inscribed in a semi-circle of radius r with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area ?
A closed cylinder has volume 2156 cm3. What will be the radius of its base so that its total surface area is minimum ?
Determine the points on the curve x2 = 4y which are nearest to the point (0,5) ?
Find the point on the parabolas x2 = 2y which is closest to the point (0,5) ?
The strength of a beam varies as the product of its breadth and square of its depth. Find the dimensions of the strongest beam which can be cut from a circular log of radius a ?
A straight line is drawn through a given point P(1,4). Determine the least value of the sum of the intercepts on the coordinate axes ?
The total area of a page is 150 cm2. The combined width of the margin at the top and bottom is 3 cm and the side 2 cm. What must be the dimensions of the page in order that the area of the printed matter may be maximum?
The space s described in time t by a particle moving in a straight line is given by S = \[t5 - 40 t^3 + 30 t^2 + 80t - 250 .\] Find the minimum value of acceleration.
Write sufficient conditions for a point x = c to be a point of local maximum.
Find the least value of f(x) = \[ax + \frac{b}{x}\], where a > 0, b > 0 and x > 0 .
For the function f(x) = \[x + \frac{1}{x}\]
The number which exceeds its square by the greatest possible quantity is _________________ .
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 point on the curve y2 = 4x which is nearest to, the point (2,1) is _______________ .
If x+y=8, then the maximum value of xy is ____________ .
If(x) = x+\[\frac{1}{x}\],x > 0, then its greatest value is _______________ .
The function f(x) = \[2 x^3 - 15 x^2 + 36x + 4\] is maximum at x = ________________ .
The maximum value of f(x) = \[\frac{x}{4 + x + x^2}\] on [ \[-\] 1,1] is ___________________ .
The sum of the surface areas of a cuboid with sides x, 2x and \[\frac{x}{3}\] and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of sphere. Also find the minimum value of the sum of their volumes.
