10.3k views
4 votes
Business/multivariable calc question
help needed asap!!!!

Business/multivariable calc question help needed asap!!!!-example-1
User MarcFasel
by
8.6k points

1 Answer

2 votes

Answer:

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

============================================================

Step-by-step 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.

User Vhbazan
by
8.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories