Answer:
Cody should run approximately 19.978 meters along the shore before jumping into the water in order to save the child.Thus,
Step-by-step explanation:
Consider the diagram below.
In this case we need to minimize the time it takes Cody to save the child.
Total time to save the child (T) = Time taken along the shore (A) + Time taken from the shore (B)
The formula to compute time is:
[tex]time=\frac{distance}{speed}[/tex]
Compute the time taken along the shore as follows:
[tex]A=\frac{x}{4}[/tex]
Compute the time taken from the shore as follows:
[tex]B=\frac{\sqrt{70^{2}+(40-x)^{2}}}{1.1}[/tex]
Then the total time taken to save the child is:
[tex]T=\frac{x}{4}+\frac{\sqrt{70^{2}+(40-x)^{2}}}{1.1}[/tex]
Differentiate T with respect to x as follows:
[tex]\frac{dT}{dx}=\frac{d}{dx}[\frac{x}{4}]+\frac{d}{dx}[\frac{\sqrt{70^{2}+(40-x)^{2}}}{1.1}][/tex]
[tex]=\frac{1}{4}-\frac{1}{1.1}\cdot \frac{(40-x)}{\sqrt{70^{2}+(40-x)^{2}}}[/tex]
Equate the derivative to 0 to compute the value of x as follows:
[tex]\frac{dT}{dx}=0[/tex]
[tex]\frac{1}{4}-\frac{1}{1.1}\cdot \frac{(40-x)}{\sqrt{70^{2}+(40-x)^{2}}}=0\\\\\frac{1}{1.1}\cdot \frac{(40-x)}{\sqrt{70^{2}+(40-x)^{2}}}=\frac{1}{4}\\\\4\cdot (40-x)=1.1\cdot [\sqrt{70^{2}+(40-x)^{2}}]\\\\\{4\cdot (40-x)\}^{2}=\{1.1\cdot [\sqrt{70^{2}+(40-x)^{2}}]\}^{2}\\\\16\cdot (40-x)^{2}=1.21\cdot [70^{2}+(40-x)^{2}}]\\\\16\cdot (40-x)^{2}-1.21\cdot (40-x)^{2}=5929\\\\14.79\cdot (40-x)^{2}=5929\\\\(40-x)^{2}=400.88\\\\40-x\approx 20.022\\\\x\approx 40-20.022\\\\x\approx 19.978[/tex]
Thus, Cody should run approximately 19.978 meters along the shore before jumping into the water in order to save the child.
