Business/multivariable calc question
help needed asap

There is a   min    value of   800   located at (x,y) =   (16, 12)  

Explanation:

Let's solve the second equation for y

4x+3y = 100

3y = 100-4x

y = (100-4x)/3

We'll plug that into the first equation

f(x,y) = 2x^2+2y^2

g(x) = 2x^2+2((100-4x)/3)^2

g(x) = 2x^2+(2/9)*(100-4x)^2

g(x) = 2x^2+(2/9)*(10,000-800x+16x^2)

This graphs a parabola that opens upward, due to the positive leading coefficient. This g(x) curve has its vertex point at the minimum.

Apply the derivative to help find the minimum

g(x) = 2x^2+(2/9)*(10,000-800x+16x^2)

g ' (x) = 4x+(2/9)*(-800+32x)

Set the derivative function equal to 0 and solve for x

g ' (x) = 0

4x+(2/9)*(-800+32x) = 0

4x+(2/9)*(-800)+(2/9)*(32x) = 0

4x-1600/9+(64/9)x = 0

9(4x-1600/9+(64/9)x) = 9*0

36x-1600+64x = 0

100x-1600 = 0

100x = 1600

x = 1600/100

x = 16

Use this x value to find y

y = (100-4x)/3

y = (100-4*16)/3

y = (100-64)/3

y = 36/3

y = 12

Therefore, (x,y) = (16,12) leads to the largest value of f(x,y) = 2x^2+2y^2

That smallest f(x,y) value is...

f(x,y) = 2x^2+2y^2

f(16,12) = 2*16^2+2*12^2

f(16,12) = 800

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.