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Kruka [31]
3 years ago
5

A cyclist travels north along a road at a constant speed of 28 miles per hour. At 1:00 P.M., a runner is 68 miles away, running

south along the same road at a constant speed. They pass each other at 3:00 P.M.. What is the speed of the runner?
Mathematics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

The speed of the runner is 6 miles per hour.

Step-by-step explanation:

Given the distance between the cyclist and the runner is 68 miles.

Given the speed of the cyclist is 28 miles per hour.

Initial time is 1:00PM Final time is 3:00PM

Time elapsed is 3-1=2hours

Let the cyclist travel a distance of x miles,Then the distance travelled by the runner is 68-x miles.

speed=\frac{distance}{time}

for the cyclist

28=\frac{x}{2}

x=56 miles

the distance travelled by the runner is 68-56=12 miles

speed of the runner =  12/2 = 6 miles per hour

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Complete question is;

Keegan is printing and selling his original design on t-shirts. He has concluded that for x shirts, in thousands sold his total profits will be p(x) = -x³ + 4x² + x dollars, in thousands will be earned. How many t-shirts (rounded to the nearest whole number) should he print in order to make maximum profits? What will his profits rounded to the nearest whole dollar be if he prints that number of shirts?

Answer:

Number of t-shirts to make maximum profit = 2790 shirts

Maximum profit = $12,209

Step-by-step explanation:

From the question, we are given that the profit function is;

p(x) = -x³ + 4x² + x

For the maximum value of the profit function,

(dp/dx) = 0 and (d²p/dx²) < 0

Since, p(x) = -x³ + 4x² + x

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-3x² + 8x + 1 = 0

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x = -0.12 or 2.79

Also, (d²p/dx²) = -6x + 8

Now, let's put the roots of x into -6x + 8 and check for maximum value conditon;

at x = -0.12

(d²p/dx²) = -6(0.12) + 8 = 7.28 > 0

At x = 2.79

(d²p/dx²) = -6(2.79) + 8 = -8.74 < 0

Maximum has to be d²p/dx² < 0

So, the one that meets the condition is -8.74 < 0 at x = 2.79

Thus, the maximum of the profit function exists when the number of shirts, x = 2.79 (in thousands) = 2790

Now, the maximum profits that corresponds to this number of t-shirts of 2.79(in thousands) is obtained by putting 2.79 for x in the profit function;

So,

p(2.79) = -(2.79)³ + 4(2.79²) + 2.79

p(x) = -21.7176 + 31.1364 + 2.79

p(x) = 12.2088 (in thousand dollars) ≈ $12,209

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