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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

$\begin{array}{1 1} \large\frac{-28\pi}{4+\pi} &\large\frac{112}{4+\pi} \\ \large\frac{28\pi}{4-\pi} &\large\frac{112}{4-\pi} \\ \large\frac{28\pi}{4+\pi} &\large\frac{112}{4+\pi} \\ \large\frac{28\pi}{4+\pi} &\large\frac{-112}{4+\pi} \end{array} $

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1 Answer

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Toolbox:
  • Area of a circle=$\pi r^2$
  • Perimeter=$2\pi r$
Step 1:
Let one part be of length $x$,then the other part =$28-x$
Let the part of the length $x$ be convered into a circle of radius.
$2\pi r=x$
$r=\large\frac{x}{2\pi}$
Area of the circle $=\pi r^2$
$\qquad\qquad\qquad=\pi\big(\large\frac {x}{2\pi}\big)^2$
$\qquad\qquad\qquad=\pi.\large\frac{x^2}{4\pi^2}$
$\qquad\qquad\qquad=\large\frac{x^2}{4\pi}$
Step 2:
Now second part of length $28-x$ is converted into square.
Side of square=$\large\frac{28-x}{4}$
Area of square $=\big[\large\frac{28-x}{4}\big]^2$
Total area $A=\large\frac{x^2}{4\pi}+\big(\frac{28-x}{4}\big)^2$
$\large\frac{dA}{dx}=\frac{2x}{4\pi}+\frac{2}{16}$$(28-x)(-1)$
$\qquad=\large\frac{x}{2\pi}-\frac{28-x}{8}$
Step 3:
$\large\frac{dA}{dx}$$=0$
$\Rightarrow \large\frac{x}{2\pi}-\frac{28-x}{8}$$=0$----(1)
$4x=28\pi-\pi x$
$4x=28\pi-\pi x$
$4x+\pi x=28\pi$
$x[4+\pi]=28\pi$
$x=\large\frac{28\pi}{4+\pi}$
Step 4:
Other part=$28-x=28-\large\frac{28\pi}{4+\pi}$
$\qquad\qquad\qquad\quad=\large\frac{112+28\pi-28\pi}{4+\pi}$
$\qquad\qquad\qquad\quad=\large\frac{112}{4+\pi}$
Step 5:
Differentiating (1)
$\large\frac{d^2A}{dx^2}=\frac{1}{2\pi}+\frac{1}{8}$=+ve
$A$ is minimum
When $x=\large\frac{28\pi}{4+\pi}$ & $28-x=\large\frac{112}{4+\pi}$
answered Aug 8, 2013 by sreemathi.v
edited Aug 8, 2013 by sreemathi.v
 

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