6. two posts, one 2.5 m high and the other 3.7 m high, stand 5 m apart. they are to be stayed by wires…

Explanation:

Step1: Set up variables

Let the distance of the stake from the (2.5)-m post be (x) m. Then the distance from the (3.7)-m post is ((5 - x)) m. The length of the wire (L) is given by (L=\sqrt{x^{2}+2.5^{2}}+\sqrt{(5 - x)^{2}+3.7^{2}}).

Step2: Differentiate (L) with respect to (x)

Using the chain - rule, if (y = \sqrt{u}), then (y^\prime=\frac{u^\prime}{2\sqrt{u}}). For (y_1=\sqrt{x^{2}+2.5^{2}}), (u_1=x^{2}+6.25), (y_1^\prime=\frac{2x}{2\sqrt{x^{2}+6.25}}=\frac{x}{\sqrt{x^{2}+6.25}}). For (y_2=\sqrt{(5 - x)^{2}+3.7^{2}}), (u_2=(5 - x)^{2}+13.69=25-10x+x^{2}+13.69=x^{2}-10x + 38.69), (y_2^\prime=\frac{2(x - 5)}{2\sqrt{x^{2}-10x + 38.69}}=\frac{x - 5}{\sqrt{x^{2}-10x + 38.69}}). (L^\prime=\frac{x}{\sqrt{x^{2}+6.25}}+\frac{x - 5}{\sqrt{x^{2}-10x + 38.69}}) To find the minimum, set (L^\prime = 0), (\frac{x}{\sqrt{x^{2}+6.25}}=-\frac{x - 5}{\sqrt{x^{2}-10x + 38.69}}) (cross - multiply) (x\sqrt{x^{2}-10x + 38.69}=-(x - 5)\sqrt{x^{2}+6.25}) Square both sides: (x^{2}(x^{2}-10x + 38.69)=(x - 5)^{2}(x^{2}+6.25)) (x^{4}-10x^{3}+38.69x^{2}=(x^{2}-10x + 25)(x^{2}+6.25)) (x^{4}-10x^{3}+38.69x^{2}=x^{4}+6.25x^{2}-10x^{3}-62.5x+25x^{2}+156.25) (x^{4}-10x^{3}+38.69x^{2}-x^{4}-6.25x^{2}+10x^{3}+62.5x - 25x^{2}-156.25 = 0) ((38.69-6.25 - 25)x^{2}+62.5x-156.25 = 0) (7.44x^{2}+62.5x-156.25 = 0) Using the quadratic formula (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) with (a = 7.44), (b = 62.5), (c=-156.25) (x=\frac{-62.5\pm\sqrt{62.5^{2}-4\times7.44\times(-156.25)}}{2\times7.44}=\frac{-62.5\pm\sqrt{3906.25 + 4650}}{14.88}=\frac{-62.5\pm\sqrt{8556.25}}{14.88}=\frac{-62.5\pm92.5}{14.88}) We take the positive root (x=\frac{- 62.5+92.5}{14.88}=\frac{30}{14.88}\approx2)

Answer:

The stake should be placed approximately (2) m from the (2.5) - m post.