# Thread: Optimisation in Two dimensions

1. ## Optimisation in Two dimensions

I have to find all the critical points and their nature from the following scenario:

There are 2 types of bread- plain and gluten free. Weekly production is represented by x hundred plain loaves and y hundred gluten free. Plain sells for 90 cents each and gluten free for $5.20 each. Weekly production costs are given by: 10x^3 + 20y^3 -10xy I have started by putting x and y as functions: fx= 30x^2 - 10y fy= 60y^2 - 10x Making them equal zero gave me x= 6y^2 y= 3x^2 How would you solve x=6(3x^2)^2 to find the critical points? What would I have to do after? Thanks 2. Originally Posted by Pandora I have to find all the critical points and their nature from the following scenario: There are 2 types of bread- plain and gluten free. Weekly production is represented by x hundred plain loaves and y hundred gluten free. Plain sells for 90 cents each and gluten free for$5.20 each. Weekly production costs are given by:
10x^3 + 20y^3 -10xy

I have started by putting x and y as functions:
fx= 30x^2 - 10y
fy= 60y^2 - 10x

Making them equal zero gave me
x= 6y^2
y= 3x^2

How would you solve x=6(3x^2)^2 to find the critical points?
What would I have to do after?

Thanks
$x = 6(3x^2)^2$

$x = 6(9x^4)$

$x = 54x^4$

$0 = 54x^4 - x$

$0 = x(54x^3 - 1)$

$0 = x[(\sqrt[3]{54}x)^3 - 1^3]$

$0 = x[(3\sqrt[3]{2}x)^3 - 1^3]$

$0 = x(3\sqrt[3]{2}x - 1)(9\sqrt[3]{4}x^2 + 3\sqrt[3]{2}x + 1 )$

So $x = 0$ or $3\sqrt[3]{2}x - 1 = 0$ or $9\sqrt[3]{4}x^2 + 3\sqrt[3]{2}x + 1 = 0$.

Solve for $x$ in each case.

3. That was very helpful, thank you! Would you mind telling me how to work out the nature of those 3 critical points also? I know you have to use the Hessian matrix, however, I am pretty terrible at partial derivatives.