Answer:
[tex]x =\dfrac{45 \sqrt{6}}{ 2}[/tex]
Step-by-step explanation:
From the given information:
The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.
Now, let V(x) be the time needed for the runner to reach the buoy;
∴ We can say that,
[tex]\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}[/tex]
In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.
i.e
The differential of V(x) = V'(x) =0
=[tex]\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0[/tex]
[tex]-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0[/tex]
[tex]\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}[/tex]
[tex]\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}[/tex]
[tex]\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}[/tex]
[tex]\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}[/tex]
squaring both sides; we get
[tex]\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}[/tex]
[tex]\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}[/tex]
By cross multiplying; we get
[tex]49x^2 = 25(54^2+x^2)[/tex]
[tex]49x^2 = 25 \times 54^2+ 25x^2[/tex]
[tex]49x^2-25x^2 = 25 \times 54^2[/tex]
[tex]24x^2 = 25 \times 54^2[/tex]
[tex]x^2 = \dfrac{25 \times 54^2}{24}[/tex]
[tex]x =\sqrt{ \dfrac{25 \times 54^2}{24}}[/tex]
[tex]x =\dfrac{5 \times 54}{\sqrt{24}}[/tex]
[tex]x =\dfrac{270}{\sqrt{4 \times 6}}[/tex]
[tex]x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}[/tex]
[tex]x =\dfrac{45 \sqrt{6}}{ 2}[/tex]
