Final answer:
The quantity to be optimized is the distance between a point on the parabola y=9-r² and the point (0,6). We can find the point on the parabola that is closest to (0,6) by finding the minimum distance. To do this, we differentiate the distance function, find critical points, and evaluate the distance function at these points to find the minimum distance. The final answer will be the point on the parabola that corresponds to the minimum distance.
Step-by-step explanation:
(a) The quantity that needs to be optimized is the distance between the point (0,6) and a point on the parabola y=9-r². We want to find the point on the parabola that is closest to (0,6).
(b) To model this situation, we can draw a graph of the parabola y=9-r² and plot the point (0,6).
(c) The distance between two points in a coordinate plane can be found using the distance formula: d = sqrt((x₂-x₁)² + (y₂-y₁)²). In this case, let x be the distance on the x-axis from the point (0,6) to the point on the parabola. The function to be optimized is the distance between the two points, which is d = sqrt(x² + (9-x²-6)²). The domain of this function is the values of x that make the expression inside the square root non-negative.
(d) To find the point on the parabola that is closest to (0,6), we differentiate the distance function and set it equal to 0 to find critical points. We then evaluate the distance function at these critical points to find the minimum distance.
(e) The final answer will be the point on the parabola that corresponds to the minimum distance.
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