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katen-ka-za [31]
1 year ago
8

The revenue function for a sporting goods company is given by R(x)=x⋅p(x) dollars where x is the number of units sold and p(x)=4

00−0.25x is the unit price. Find the maximum revenue.
Mathematics
1 answer:
Zina [86]1 year ago
5 0

The maximum revenue generated is $160000.

Given that, the revenue function for a sporting goods company is given by R(x) = x⋅p(x) dollars where x is the number of units sold and p(x) = 400−0.25x is the unit price. And we have to find the maximum revenue. Let's proceed to solve this question.

R(x) = x⋅p(x)

And,  p(x) = 400−0.25x

Put the value of p(x) in R(x), we get

R(x) = x(400−0.25x)

R(x) = 400x - 0.25x²

This is the equation for a parabola.  The maximum can be found at the vertex of the parabola using the formula:

x = -b/2a from the parabolic equation ax²+bx+c   where a = -0.25,  b = 400 for this case.

Now, calculating the value of x, we get

x = -(400)/2×-0.25

x = 400/0.5

x = 4000/5

x = 800

The value of x comes out to be 800. Now, we will be calculating the revenue at x = 800 and it will be the maximum one.

R(800) = 400x - 0.25x²

= 400×800 - 0.25(800)²

= 320000 - 160000

= 160000

Therefore, the maximum revenue generated is $160000.

Hence, $160000 is the required answer.

Learn more in depth about revenue function problems at brainly.com/question/25623677

#SPJ1

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Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

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Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

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