Business/multivariable calc question help needed asap

I solved and got a max of (8/5,8) at 64/5

Answers:

There is a  max   value of   81/8   located at (x,y) =   (9/8, 9)  

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Explanation:

Solve the second equation for y

8x+y = 18

y = 18-8x

Plug this into the first equation

f(x,y) = x*y

g(x) = x*(18-8x)

g(x) = 18x-8x^2

This graphs out a parabola that opens downward, and has a max point at the vertex.

If you apply the derivative to this, you get g ' (x) = 18 - 16x

Set this equal to zero and solve for x

g ' (x) = 0

18 - 16x = 0

18 = 16x

16x = 18

x = 18/16

x = 9/8

Use this x value to find y

y = 18 - 8x

y = 18 - 8(9/8)

y = 18 - 9

y = 9

So the max is x*y = (9/8)*9 = 81/8

Or we could say

g(x) = 18x-8x^2

g(9/8) = 18(9/8)-8(9/8)^2

g(9/8) = 81/8

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To summarize,

There is a  max   value of   81/8   located at (x,y) =   (9/8, 9)  

When saying "max value of something", we're basically talking about the largest f(x,y) value. Which in this case is the largest x*y value based on the fact that 8x+y = 18.

A practical real world example of a problem like this would be if you wanted to max out a certain rectangular area based on constraints of how much building material you have for the fence.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

found it xd it ok yt