Let a piece of length l be cut from the given wire to make a square.
Then,the other piece of wire to be made into a circle is of length\((28−l) m.\)
Now,side of square =\(\frac{l}{4}\).
Let r be the radius of the circle.Then,\(2\pi r=28-l\)⇒\(r=\frac{1}{2\pi}(28-l).\)
The combined areas of the square and the circle\((A)\)is given by,
\(A=(side \space of\space the \space square)\pi ^{{2}}+r^{2}\)
\(=\frac{l^{2}}{16}+\pi[\frac{1}{2\pi}(28-l)]^{2}\)
\(=\frac{l^{2}}{16}+\frac{1}{4\pi}(28-l)^{2}\)
\(∴\frac{dA}{dl}\)=\(\frac{2l}{16}+\frac{2}{4\pi }(28-l)(-1)\)\(=\frac{l}{8}-\frac{1}{2\pi}(28-l)\)
\(\frac{d^{2}A}{dl^{2}}=\frac{1}{8}+\frac{1}{2\pi }>0\)
Now,\(\frac{dA}{dl}=0⇒\frac{l}{8}-\frac{1}{2\pi}(28-l)=0\)
\(⇒\frac{\pi l-4(28-l)}{8\pi}=0\)
\(⇒(\pi +4)l-112=0\)
\(⇒l=\frac{112}{\pi+4}\)
Thus,when\(⇒l=\frac{112}{\pi+4}\),\(\frac{d^{2}A}{dl^{2}}>0\).
By second derivative test,the area(A)is the minimum when \(l=\frac{112}{\pi+4}\)
Hence,the combined area is the minimum when the length of the wire in making the
square is \(l=\frac{112}{\pi+4}\) cm while the length of the wire in making the circle
is \(28-\frac{112}{\pi+4}=\frac{28\pi}{\pi+4} cm.\)
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Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is