Business/multivariable calc question
help needed asap

There is a   min     value of   376     located at (x,y) =   (9,7)  

Explanation:

Solve the second equation for y

x+y = 16

y = 16-x

Then plug it into the first equation

f(x,y) = 3x^2+4y^2 - xy

g(x) = 3x^2+4(16-x)^2 - x(16-x)

g(x) = 3x^2+4(256 - 32x + x^2) - 16x + x^2

g(x) = 3x^2+1024 - 128x + 4x^2 - 16x + x^2

g(x) = 8x^2-144x+1024

The positive leading coefficient 8 tells us we have a parabola that opens upward, and produces a minimum value (aka lowest point) at the vertex.

Let's compute the derivative and set it equal to zero to solve for x.

g(x) = 8x^2-144x+1024

g ' (x) = 16x-144

16x-144 = 0

16x = 144

x = 144/16

x = 9

The min value occurs when x = 9. Let's find its paired y value.

y = 16-x

y = 16-9

y = 7

The min value occurs at (x,y) = (9,7)

Lastly, let's find the actual min value of f(x,y).

f(x,y) = 3x^2+4y^2 - xy

f(9,7) = 3(9)^2+4(7)^2 - 9*7

f(9,7) = 376

The smallest f(x,y) value is 376.

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: