Can someone interpret the question please? I’m my understanding. Thank you ☺️
1 answer:
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Answer:
Santos should run approximately 36.145 meters along the shore
Step-by-step explanation:
The given parameters are;
The distance along the shore of the child = 50 meters
The distance from the shore of the child = 60 meters
The rate at which Santos can run = 4 m/s
The rate he can swim = 0.9 m/s
Let 'x' represent the distance he runs along the shore
We have;
The time he spends running on the shore, t₁ = x/4
The time he spends swimming, t₂ = (√(60² + (50 - x)²)/0.9
The total time, T = t₁ + t₂ = x/4 + (√(60² + (50 - x)²)/0.9
To find the maximum, we have;
dT/dx = 0 = d(x/4 + (√(60² + (50 - x)²)/0.9)/dx = 1/4 - (50 - x)/(0.9·(√(60² + (50 - x)²) = 0
1/4 - (50 - x)/(0.9·(√(60² + (50 - x)²) = 0
Simplifying using a graphing calculator gives;
1519·x² - 151900·x + 3505900 = 0
x = (151900 ± √((-151900)² - 4×1519×3505900))/(2 × 1519)
x ≈ 63.86 m or x ≈ 36.145 m
We note that the distance from a point x = 63.83 meters and 36.145 meters from where Santos spots the girl to the location of the girl are the same
Therefore, Santos should run approximately 36.145 meters along the shore before jumping into to the water in order to save the child
Answer:
26 ft square by 13 ft high
Step-by-step explanation:
The tank will have minimum surface area when opposite sides have the same total area as the square bottom. That is, their height is half their width. This makes the tank half a cube. Said cube would have a volume of ...
2·(8788 ft^3) = (26 ft)^3
The square bottom of the tank is 26 ft square, and its height is 13 ft.
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<em>Solution using derivatives</em>
If x is the side length of the square bottom, the height is 8788/x^2 and the area is ...
x^2 + 4x(8788/x^2) = x^2 +35152/x
The derivative of this is zero when area is minimized:
2x -35152/x^2 = 0
x^3 = 17576 = 26^3 . . . . . multiply by x^2/2, add 17576
x = 26
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As the attached graph shows, a graphing calculator can also provide the solution.
