Here we have , A manufacturer finds that the revenue generated by selling of cortan commodity is given by function R(x)=80x-0.2x^ 2 , where he maximum reveremany should be manufactured to obtain this maximum units .Let's find out:

We have following function as $R(x)=80x-0.2x^ 2$ . Let's differentiate this and equate it to zero to find value of x for which the function is maximum!

⇒ $R(x)=80x-0.2x^ 2$

⇒ $\frac{d(R(x))}{dx}=\frac{d(80x-0.2x^ 2)}{dx}$

⇒ $\frac{d(R(x))}{dx}=\frac{d(80x)}{dx}-\frac{d(0.2x^ 2)}{dx}}$

⇒ $0=80-2x(0.2)$

⇒ $\frac{80}{0.4}=x$

⇒ $x=200$

Now , Value of function at x=200 is :

⇒ $R(200)=80(200)-0.2(200)^ 2$

⇒ $R(200)=16000-8000$

⇒ $R(200)=8000$

Therefore , The maximum units is 200 and , Total revenue is $8,000

7 0

3 years ago

A hiker climbs a mountain. After 1 hour he is at an altitude of 200 feet. After 3 hours, he is at an altitude of 350 feet. What

s344n2d4d5 [400]

Answer: 75 feet per hour

Step-by-step explanation:

X: 1, 3

Y: 200, 350

150/2

6 0

3 years ago

The sequence a1 = 6, an = 3an-1 can also be written as

wariber [46]

The question as presented is incomplete, here is the complete question with the multiple choice:

The sequence a1 = 6, an = 3an − 1 can also be written as:

1) an = 6 ⋅ 3^n
2) an = 6 ⋅ 3^(n + 1)
3) an = 2 ⋅ 3^n
4) an = 2 ⋅ 3^(n + 1)

The correct choice is option 3) an = 2⋅3^n.

If we look at the initial sequence an = 3⋅an-1, and

a1 = 3⋅a0 = 6
a0 = 6/3
a0 = 2

We can now look at the sequence.

a0 = 2
a1 = 6
a2 = 18
a3 = 54
etc...

A common factor in each of those numbers is 2, so we can rewrite the sequence by factoring out 2.

a0 = 2⋅1
a1 = 2⋅3
a2 = 2⋅9
a3 = 2⋅27

The numbers being multiplied by 2 are all factors of 3. So we can rewrite the sequence again as:

a0 = 2⋅3^0
a1 = 2⋅3^1
a2 = 2⋅3^2
a3 = 2⋅3^3

This sequence can now be rewritten as an = 2⋅3^n.

7 0

3 years ago