See the picture Please help

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

Zina [86]

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

5 0

1 year ago

In need of some help please.

Hatshy [7]

By the Pythagorean theorem
50^2 = x^2 +(x +34)^2
2500 = 2x^2 +68x + 1156
x^2 +34x -672 = 0
(x -14)(x +48) = 0
x = 14 or -48

The distance from the wall to the base of the ladder is 14 ft.

_____
7-24-25 is a Pythagorean triple. The dimensions here are double those values. It can be handy to know a few of the Pythagorean triples, as that can let you write down the answers to problems without having to go through the equations.

8 0

3 years ago

11. Given a matrix A, what is A^1? A=[0 12 0 1]

pogonyaev

$\text{Since det(A) = 0, the inverse of the matrix does not exist.}$

3 0

2 years ago

If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.

Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have $S_1=3\times2^{1-1}+2(-1)^1=3-2=1$ , which agrees with the initial value S₁ = 1.

• Assume the closed form is correct for all n up to n = k. In particular, we assume

$S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}$

and

$S_k=3\times2^{k-1}+2(-1)^k$

We want to then use this assumption to show the closed form is correct for n = k + 1, or

$S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}$

From the given recurrence, we know

$S_{k+1}=S_k+2S_{k-1}$

so that

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)$

$S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}$

$S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)$

$S_{k+1}=3\times2^k-2(-1)^k$

$S_{k+1}=3\times2^k+2(-1)(-1)^k$

$\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}$

which is what we needed. QED

6 0

2 years ago

Find the domain and range of the graph

ozzi

9514 1404 393

Answer:

domain: [-6, -1)
range: [-5, 4]

Step-by-step explanation:

You have correctly determined the domain to be from x = -6 to x < -1. In interval notation, that would be ...

[-6, -1)

__

The range is the vertical extent of the graph. It extends from a minimum of y = -5 to a maximum of y = +4 at the top of the curve. In interval notation, that would be ...

[-5, 4]

8 0

2 years ago