Consider the following problem. Find two numbers whose sum is 23 and whose produ

Question

Consider the following problem. Find two numbers whose sum is 23 and whose product is a maximum. a) Make a table of values, like the following one, so that the sum of the numbers in the first two columns is always 23. On the basis of the evidence in your table, estimate the answer to the problem. What is the maximum vertical distance between the line y = x + 2 and the parabola y = x^2 for - 1 lessthanorequalto x lessthanorequalto 2? Find the point on the line y = 2x + 3 that is closest to the origin. Find the points on the ellipse 4 x^2 + y^2 = 4 that are farthest away from the point (1, 0). Find the most general antiderivative of the function (Check your answer by differentiation). f(x) = 1/2 x^2 - 2x + 6 f(x) = x(2 - x)^2 g(t) = 1 + t + t^2/squareroot t f(x) = 2 squareroot x + 6 cos x

Explanation / Answer

Solution:(19)

y = 2x + 3
m = 2
Perpendicular line:
m = –(1/2)
Equation of perpendicular line parring through origin.
(y – 0) = –(1/2)(x – 0)
y = –(1/2)x
2y = –x
x + 2y = 0
Find intersection with given line:
x + 2(2x + 3) = 0
x + 4x + 6 = 0
5x = –6
x = –6/5
y = 2(–6/5) + 4 = –12/5 + 4 = 8/5
(x, y) = (–6/5, 8/5) .... point on y = 2x + 3 closest to origin

Solution:(21)

Let R be the distance from (1,0) to a point on the ellipse:

(x-1)2 + y2 = R2 ==> y2 = R2 - (x-1)2

4x2 + y2 = 4 ==> y2 = 4 - 4x2

R2 - (x-1)2 = 4 - 4x2
R2 = x2 - 2x + 1 + 4 - 4x2
R2 = 5 - 2x - 3x2

Maximizing R2:

d(R2)/dx = - 2 - 6x ==> 0 ==> x = -1/3

4(-1/3)2 + y2 = 4
y2 = 4 - 4/9 = 32/9
y = ±42/3

The two points on the ellipse the farthest from (1,0) are (-1/3, -42/3) and (-1/3, 42/3)

That distance is [(-1/3 - 1)2 + (42/3 - 0)2 = (16/9 + 32/9) = (48/9) = 43/3 2.309

The distance to the vertices (0, ±2) is (12 + 22) = 5 2.236 which is less than the maximum

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