Cal 1 Optimization Question?

Question

Let Q = (0,6) and R= (8,11) be given points in the plane. We want to find the point P=(x,0) on the x-axis such that the sum of distances PQ+PR is as small as possible. (Before proceeding with this problem, draw a picture!)
To solve this problem, we need to minimize the following function of x:
f(x)=
over the closed interval [a,b] where
a=
b=
We find that f(x) has only one critical number in the interval at x=

where f(x) has value of:

Since this is smaller than the values of f(x) at the two endpoints, we conclude that this is the minimal sum of distances.

Answer 1

f(x) = √(36 + p^2) + √[121 + (8 - p)^2]
the interval is: [0, 8]
Take the derivative and set it equal to zero:
(1/2)(36 + p^2)^(-1/2)+(1/2)(p^2 - 16p + 185)^(-1/2)(p - 8) = 0
P/√(36 + p^2) + (p - 8)/√(p^2 - 16p + 185) = 0
p√(p^2 - 16p + 185) = (8 - p)√(36 + p^2)
p^2(p^2 - 16p + 185) = (8 - p)^2(36 + p^2)
p^4 - 16p^3 + 185p^2 = p^4 - 16 p^3 + 100p^2 - 576p + 2304
85p^2 + 576p - 2304 = 0
p = [-576 +/- √(576^2 - 4(85)(-2304)]/170
p = (-576 +/- 1056)/170
p = 2.8235, -9.6[extraneous]

Note that, as expected, the angles that the segments makes with horizontal are equal:
arc tan(6/2.8235) = arc tan(11/ (8 - 2.8235)) = 64.8deg

Answer 2

think PQ=d, then the area between Q and the island is sqrt(d^2+nine), the completed length of the pipe would be sqrt(d^2+nine)+(10-d) and the completed fee would be 2.4sqrt(d^2+nine)+(10-d) we would desire to decrease the above, so we would desire to consistently take the 1st by-product, yielding: 2.4*0.5*(d^2+nine)^(-0.5)*2*d-one million enable the above equivalent to 0, and resolve for the equation....finally we've d (the area between P and Q) = one million.375 miles