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LiRa [457]
3 years ago
8

To rent a certain meeting room, a college charges a reservation fee of $37 and an additional fee of $6.70 per hour. the chemistr

y club wants to spend less than $90.60 on renting the meeting room. what are the possible amounts of time for which they could rent the meeting room? use t for the number of hours the meeting room is rented, and solve your inequality for t .
Mathematics
1 answer:
mojhsa [17]3 years ago
3 0
37 + 6.70t < 90.60
6.70t < 90.60 - 37
6.70t < 53.6
t < 53.6 / 6.70
t < 8 <=== they could rent the meeting room for less then 8 hrs
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Answer:

a) The function is: f(x, y) = x + y.

The constraint is: x*y = 196.

Remember that we must write the constraint as:

g(x, y) = x*y - 196 = 0

Then we have:

L(x, y, λ) = f(x, y) +  λ*g(x, y)

L(x, y,  λ) = x + y +  λ*(x*y - 196)

Now, let's compute the partial derivations, those must be zero.

dL/dx =  λ*y + 1

dL/dy =  λ*x + 1

dL/dλ = (x*y - 196)

Those must be equal to zero, then we have a system of equations:

λ*y + 1 = 0

λ*x + 1 = 0

(x*y - 196) = 0

Let's solve this, in the first equation we can isolate  λ to get:

λ = -1/y

Now we can replace this in the second equation and get;

-x/y + 1 = 0

Now let's isolate x.

x = y

Now we can replace this in the last equation, and we will get:

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b) Now we have:

f(x) = x*y

g(x) = x + y - 196

Let's do the same as before:

L(x, y, λ) = f(x, y) +  λ*g(x, y)

L(x, y, λ) = x*y +  λ*(x + y - 196)

Now let's do the derivations:

dL/dx = y + λ

dL/dy = x + λ

dL/dλ = x + y - 196

Now we have the system of equations:

y + λ = 0

x + λ = 0

x + y - 196 = 0

To solve it, we can isolate lambda in the first equation to get:

λ = -y

Now we can replace this in the second equation:

x - y = 0

Now we can isolate x:

x = y

now we can replace that in the last equation

y + y - 196 = 0

2*y - 196 = 0

2*y = 196

y = 196/2 = 98

The maximum will be:

x*y = y*y = 98*98 = 9,604

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Answer:

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Step-by-step explanation:

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The population of interest in this study is UCB undergraduates.

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