Question

Cody is a lifeguard and spots a drowning child 40 meters along the shore and 70 meters from the shore to the child. Cody runs along the shore for a while and then jumps into the water and swims from there directly to the child. Cody can run at a rate of 4 meters per second and swim at a rate of 1.1 meters per second. How far along the shore should Cody run before jumping into the water in order to save the child? Round your answer to three decimal places.

Answer 1

Cody should run approximately 19.978 meters along the shore before jumping into the water in order to save the child.Thus,

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:

`time=(distance)/(speed)`

Compute the time taken along the shore as follows:

`A=(x)/(4)`

Compute the time taken from the shore as follows:

`B=\frac{\sqrt{70^(2)+(40-x)^(2)}}{1.1}`

Then the total time taken to save the child is:

`T=(x)/(4)+\frac{\sqrt{70^(2)+(40-x)^(2)}}{1.1}`

Differentiate T with respect to x as follows:

`(dT)/(dx)=(d)/(dx)[(x)/(4)]+(d)/(dx)[\frac{\sqrt{70^(2)+(40-x)^(2)}}{1.1}]`

`=(1)/(4)-(1)/(1.1)\cdot \frac{(40-x)}{\sqrt{70^(2)+(40-x)^(2)}}`

Equate the derivative to 0 to compute the value of x as follows:

`(dT)/(dx)=0`

`(1)/(4)-(1)/(1.1)\cdot \frac{(40-x)}{\sqrt{70^(2)+(40-x)^(2)}}=0\\\\(1)/(1.1)\cdot \frac{(40-x)}{\sqrt{70^(2)+(40-x)^(2)}}=(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`

Thus, Cody should run approximately 19.978 meters along the shore before jumping into the water in order to save the child.