Calculus Optimization question.?

Question

A boat leaves a dock at 2:00pm and travels due south at a speed of 20 km/h Another boat has been heading due east at 15 km/h and reaches the same dock at 3:00pm. At what time were the 2 boats closest together?




I have no idea how to start this problem. Any help would be great. Thanks.

Answer 1

In problems like this, it's always a good idea to draw a diagram. Call the dock point D. Put a point below D to represent the first boat. Draw an arrow from the dock to that point to represent the path of the boat heading south. The distance of the boat from the dock (the length of the arrow) will be y, and you know that y' is 20. Put a point to the west of the dock to represent the second boat Draw an arrow from that point to D to represent the path of the second boat. Call x the distance of the second boat from the dock - you know that x' is 15. A line connecting the two boats will represent the distance between them - call it s. You're looking to minimize s wrt time. Note that s, x and y form a right triangle.

Does that help get you started?

Answer 2

Are you sure this is the whole word problem?
It sounds as if a piece of information is missing.

If this is univariate calculus as opposed to multivariate calculus, then it seems as if this problem above is more of a related rates problem where you are ask to minimized the distance (i.e. when the two boats are closest to each other).

Your distance equation is distance (d)== rate (r) * time (t). Take the derivative of the equation with respect to r and t. Don't forget to use the Product Rule.

This is to get you started. As I stated before, I feel that a piece of information is missing from the problem. Good luck!!

Answer 3

a million) It is realistic; ==> First make a decision the target of the hindrance; right here it's paper dimensions, in order that its discipline will probably be minimal; then subsequent comes underneath what constraints this should be performed? ==> Well, right here On the chosen paper, you're leaving a few margins in every single place and arriving on the discipline for printing, that's given a few constant importance. two) Now, we're transparent what we wish; How to continue for constructing mathematical equations, in order that it may be solved for purchasing the top reply: three) Here we're given the printing discipline as 50 squarewhere is consistent; as a result allow us to begin from this information; ==> Let the measurement of the printing discipline be x in (top smart) via y in (width smart) ==> Printing discipline = xy = 50; == y = 50/x -------(a million) four) There is a margin of four in each and every at best and backside are offered; as a result total top of the paper is "x + eight" in; Similarly a margin of two in is offered on both aspects; ==> Overall width = y + four in five) Hence the discipline of the paper is = (x+eight)(y+four) 6) Substituting for y from equation (a million), A (x+eight)(50/x + four) = 50 + four hundred/x + 4x + 32 7) Hence the position to be minimized is: A = eighty two + 4x + four hundred/x eight) Differentiating this, A' = zero + four - 2 hundred/x^two = four - four hundred/x^two nine) Equating A' = zero, x^two = one hundred; ==> x = +/- 10 in 10) But a measurement can not be bad, as a result we recollect best + 10; So x = 10 in and y = five in eleven) However we must confirm,whether or not that is minimal or highest; for which we can practice 2d spinoff experiment; So once more differentiating, A'' = 800/x^three; at x = 10, A'' is > zero; as a result it's minimal Thus we finish for the printing the discipline to be 50 squarein, underneath the given constraints of margins, the outer dimension of the paper have to be 18 in via nine in; So outer discipline is = 162 squarein. Wish you're defined; have a first-class time.

Answer 4

You can find my solution here

http://img180.imageshack.us/img180/950/boats1ui8.png