Tall, Dark, and Mysterious


Not from the solutions manual of Calculus With Analytic Geometry

File under: This One Time, At Mathcamp, Queen of Sciences. Posted by Moebius Stripper at 1:46 pm.
  1. You’re in the middle of the desert, 2 km north of an east-west river. Your destination is 1 km north and 5 km east of where you stand. You need to stop at the river for a drink at some point in your journey, but it doesn’t matter when. What is the shortest path you can take?

    Solution. Let the river be represented by the x-axis, your starting point be P(0,2), and your destination be at Q(5,3). The shortest path between P and Q that touches the x-axis is equivalent to the shortest path from P to Q’(5,-3). This path is clearly a straight line; properties of similar triangles show that you arrive at the river at (2,0).

  2. A statue is 4 m tall and mounted on a 2 m pedestal. Your eye level is 1.5 m above ground. Where should you stand in order to get the view of the statue deemed “best” by the writers of calculus textbooks - that is, the view such that the angle subtended by the statue at your eye is a maximum?

    Solution. Because I’m lazy and don’t feel like creating pictures, let’s say that the base of the pedestal is at O(0,0), the base of the statue is at B(0,2), and the top of the statue is at T(0,6). Then your eye will be somewhere on the line y=1.5. Construct the circle tangent to y=1.5 that has the segment BT as a chord. Then the point of tangency, P, is such that the angle BPT lies in the circle, while BQT lies outside the circle for all other Q on the line y=1.5. Hence BPT>BQT for all such Q, so you should stand at the P. The Pythagorean Theorem quickly shows that you should stand 1.5 m from the base of the statue.

  3. When widgets are priced at $50 apiece, 100 people buy them. For every $5 increase in price, 2 fewer people buy widgets. How much should a widget cost in order to maximize the widget company’s revenue?

    Solution. Letting x be the number of $5 price increases, the company’s revenue, which we wish to maximize, is R=(50+5x)(100-2x). Equivalently, we may maximize 5/2*R=(50+5x)(250-5x). Applying the AM-GM inequality to this quantity, we have sqrt(5/2*R) < = 1/2*((50+5x)+(250-5x))=150. The revenue is thus maximized when 50+5x=250-5x, ie, x=20. So, twenty $5 price increases, for a price of $150 per widget, gives the maxmimum revenue.

  4. Find the largest possible perimeter of a rectangle whose sides are parallel to the axes and that can be inscribed in an ellipse with equation x^2+2y^2=1.

    Solution. Letting (x,y) by a point on the ellipse such that the rectangle’s vertices are at (x,y), (-x,y), (x,-y), and (-x,-y), we have a perimeter of P=4x+4y to maximize. Apply Cauchy-Schwartz to the vectors (x, sqrt(2)y) and (1,1/sqrt(2)) to obtain P/4=x+y< =sqrt(x^2+2y^2)sqrt(1+1/2)=sqrt(3/2). The perimeter is a maximum when x=2y. Substituting into the equation of the ellipse, x=2sqrt(6), y=sqrt(6), and so the largest perimeter possible is 12sqrt(6).

  5. Two hallways meet one another at right angles. One hallway is 27 inches wide, and the other is 64 inches wide. What is the length of the longest ladder that can be carried horizontally around the corner?

    Solution. The ladder’s length, L, is that of the shortest line segment that passes through the point (64, -27) and is bounded by the axes. The lines passing through (64,27) have equations of the form y=mx-64m-27. The intercepts of such a line are at (0,(-64m-27)) and ((64m+27)/m,0), so we need to minimize L^2=(64m+27)^2+((64m+27)^2)/m^2=(1/m^2+1)(64m+27)^2. Apply Holder’s Inequality with three 3-norms to the vectors (1/m^(2/3),1), (4m^(1/3), 3) and (4m^(1/3), 3). This gives L^(2/3) = ((1/m^2+1)(64m+27)^2)^(1/3) >= (64)^(2/3)+27^(2/3), that is, L>=25^(3/2)=125. So the longest ladder that can be carried horizontally around the corner is 125 inches long.