Question

Mason is standing on the seashore. He believes that if he makes a wish

and throws a seashell back into the ocean, his wish will come true. Mason is
standing at the origin of a coordinate plane and the shoreline is represented by the
graph of the line
y = 1.5x + 13. Each unit represents 1 meter. How far does Mason need to be able
to throw the seashell to throw one into the ocean? Round your answer to the
nearest centimeter.

Answer 1

To throw a seashell into the ocean, Mason needs to throw it above the line y = 1.5x + 13, which represents the shoreline.

This means that the seashell's height, y, must be greater than 1.5 times its horizontal distance, x, plus 13. We can use the Pythagorean theorem to find the distance, d, that Mason needs to throw the seashell, given by d = sqrt(x^2 + y^2).

We want to find the minimum value of d that satisfies y > 1.5x + 13. This occurs when y is equal to 1.5x + 13, since any larger value of y would require a larger value of d.

So we can substitute y = 1.5x + 13 into the equation for d and get:

d = sqrt(x^2 + (1.5x + 13)^2)

To find the minimum value of d, we can use calculus and find the derivative of d with respect to x, and set it equal to zero. Alternatively, we can use a graphing calculator or an online tool to plot the function d and find its minimum point. Either way, we get that the minimum value of d occurs when x is approximately -5.2 meters and y is approximately 5.2 meters. The corresponding value of d is approximately 14.7 meters.

Therefore, Mason needs to be able to throw the seashell at least 14.7 meters, or 1470 centimeters, to throw it into the ocean.