Optimization Problems II

A long rectangular sheet of metal, 12 inches wide, is to be made into a rain gutter by turning up two sides so that they are perpendicular to the sheet. How many inches should be turned up to give the gutter its greatest capacity?
1. We drew the diagram and assigned the variables, base (b) and height (h). We realized that the b=12-2h, since the rain gutter takes the shape of an open rectangle with only two sides.

2. We also realized that this is an area problem, not a volume problem.

A=b*h
A=(12-2h)h..........The definition of the base is given in terms of height. From the equation that we set up, we can see that the area is a function of height.
A(h)=12+2h^2
A'(h)=12-4h.........Differentiated the previous equation so that we may find the critical numbers by finding the x-intercepts of the derivative of A.
0 = 12-4h
h = 3......................Determine if the critical numbers are a maximum or a minimum using the 1st derivative test.

3. We then determined the domain of which the function can exist using the context in the problem. The lowest possible input of the graph would be 0, which would give us an area of 0 (imagine two walls stuck together). The highest possible input of the graph would be 6, which would also give us an area of 0 (imagine two walls stuck together).

Domain: [0, 6]

4. We realized that to the left of h=3 on the graph of A', the function is positive, and to the right, negative. We know this by plugging a number <3>3 into the equation of A'. Therefore, h=3 is a local max. This is the first derivative test.

Answer: 3 inches


A North-South highway intersects an East-West highway at a point P. A car crosses P at 10:00 a.m., traveling east at a constant speed of 20 mph. At that same instant another car is 2 miles north of P, traveling south at 50 mph. Find the time at which they are closest to each other and approximate the minimum distance between the automobiles.

1. Drew the diagram and laid out the info.

• Point P is the intersection.
• Car A is moving 20mph west.
• Car B is moving 50mph south, north of P.
• We can write designate the distance as a function of time. The derivative of the distance function is the velocity function. So (dB)/(dt) = 50 and (dA)/(dt) = 20.
• Designated the hypotenuse of the triangle, d.

2. Using the info provided, we organized the info onto a chart, where d=distance, r=velocity (a.k.a. rate), and t=time of cars A and B.

Using the chart, we were able to generate a formula for d, the distance or the hypotenuse or the hypotenusal distance. Yes, that's my adjective for hypotenuse; it's been copyrighted by none other than me, zeph.

3. Because d is a radical function and we're looking for the critical numbers, uh oh, I forgot. (Someone, help me out here.) Then we expand the equation we generated for d, so we can apply the derivative rules as minimum as possible. (Who wants to use chain rule 20 times for deriving an equation? I don't.) We did this so we can find the critical number(s), determine if the critical number(s) is a max or min using the 1st derivative test, then solve for t.

Answer: t=1/29 hr, which is approx, 2 min, so at 10:02am the cars are closest.

4. Since we know t=1/29 hr, we put that into d to determine the minimum distance at 10:02am.

Answer: d(1/29)=0.7428 miles.


END NOTES

• Homework: 5.4 #7-12
• Slide #7 has a question for you to solve.
  
• We will be continuing optimization problems on Monday.
• Passing the baton to Joyce.