A boat on the ocean is 4 mi from the nearest point on a straight shoreline; that point is 6 mi from a restaurant on the shore. A woman plans to row the boat straight to a point on the shore and then walk along the shore to the restaurant. a. If she walks at 3 mi/hr and rows at 2 mi/hr, at which point on the shore should she land to minimize the total travel time? b. If she walks at 3 mi/hr, what is the minimum speed at which she must row so that the quickest way to the restaurant is to row directly (with no walking)?

Question

A boat on the ocean is 4 mi from the nearest point on a straight shoreline; that point is 6 mi from a restaurant on the shore. A woman plans to row the boat straight to a point on the shore and then walk along the shore to the restaurant. a. If she walks at 3 mi/hr and rows at 2 mi/hr, at which point on the shore should she land to minimize the total travel time?   b. If she walks at 3 mi/hr, what is the minimum speed at which she must row so that the quickest way to the restaurant is to row directly (with no walking)?

Unpled · Accepted Answer

a) On the ocean, we have a boat that is&amp;nbsp;4mi&amp;nbsp;on a straight stretch of shoreline, and we are aware that this point is&amp;nbsp;6mi&amp;nbsp;distance from a waterfront restaurant. The activity begins with a &amp;nbsp;3mihr&amp;nbsp;and rows at&amp;nbsp;2mihr&amp;nbsp;and we must determine at which point on the shore she should land to minimize the total travel time.&amp;nbsp;Let x represent the separation between the landing location and the closest coastline point. According to this, the distance between her landing spot and the restaurant is equivalent to (6 - x)mi.Let t stand in for the amount of time it took her to go to the restaurant. It would be beneficial to represent t as a function of x (t = t(x)) and then equalize its derivative with zero in order to perform this minimization.The distance between the boat and its landing is equal to thanks to the Pythagoras Theorem.d=42+x2=16+x2.&amp;nbsp;t(x)=16+x22+6−x3hours&amp;nbsp;For time to be minimal we place&amp;nbsp;&amp;nbsp;dt&amp;nbsp;&amp;nbsp;dx&amp;nbsp;=0, which gives us:&amp;nbsp;t‘(x)=2x416+x2−13=x216+x2−13=0&amp;nbsp;3x=216+x2&amp;nbsp;and squaring both sides we get&amp;nbsp;9x2=4(16+x2)&amp;nbsp;or&amp;nbsp;5x2=64&amp;nbsp;Since we are discussing distance, we must select the positive of the two solutions to this equation, which isx=85mi=858mi.&amp;nbsp;So, if she lands at the poin that is&amp;nbsp;6−855mi=30−855mi&amp;nbsp;away from the restaurant, she will minimie the time needed to get to the restaurant.

memomzungup4 · Accepted Answer

b) We now know that she walks at 3mihr and we must determine the minimum speed at which she must row so that the quickest way to the restaurant is to row directly (without walking). Since now we don&#039;t have the speed of rowing, the time will be: t(x)=16+x2vr+6−x3 Let&#039;s now determine vr from here and later on incorporate the condition that we have for x. 16+x2vr=t−6−x3=3t−t+x3 vr=316+x23t−6+x For vr to be minimal we have that dvrdx=0. dvrdx=3. 2x216+x2⋅(3t−6+x)−16+x2⋅1(3t−6+x)2=0, which implies that: 2x216+x2⋅(3t−6+x)−16+x2⋅1=0 3tx−6x+x2−16−x216+x2=0, which imlies that 3tx−6x+x2−16−x2=0ort=16+6x3x (1) Since we want to find the minimum speed of reaching the restaurant without walking, we have that x = 6, so if we incorporate that into equation (1) we get that t=16+3618=5218hr. We now return vr and by incorporating t=5218hr we get that vr=16+625218=525218=91313mihr We conclude that the minimal speed at which she must row so that the quickest way to the restaurant is without walking is vr=91313mihr

A boat on the ocean is 4 mi from the nearest point on a straight shore

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