The path of his motorcycle was given approximately by H=-0.005x^2+2.39x+600 where H was measured in ft above the river and x was the distance from his launch ramp. How high above the river was the launch ramp? I found that the maximum height was 600ft and the horizontal distance was 234ft.

The path of a motorcycle is given by [tex]H =-0.005x^{2}+2.39x + 600[/tex] .....(1) where H is the height above the river in ft and x is the distance from his launch camp.

Putting x = 0, the height of the launch camp from the river is H = 600 ft.

Now. differentiating equation (1), with respect to x both sides we get,

[tex]\frac{dH}{dx} = 0.01x -2.39 =0[/tex] {Condition for H to be maximum is [tex]\frac{dH}{dx} =0[/tex]}

⇒ x = 239 ft.

So, [tex]H_{max} = -0.005 (239)^{2} + 2.39 \times 239 + 600 =885.6[/tex] ft. (Answer)

The path of his motorcycle was given approximately by H=-0.005x^2+2.39x+600 where H was measured in ft above the river and x was the distance from his launch ramp. How high above the river was the launch ramp? I found that the maximum height was 600ft and the horizontal distance was 234ft.​

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The path of his motorcycle was given approximately by H=-0.005x^2+2.39x+600 where H was measured in ft above the river and x was the distance from his launch ramp. How high above the river was the launch ramp? I found that the maximum height was 600ft and the horizontal distance was 234ft.