use Lagrange Multipliers to find max & min of f(x,y)=5xy constrained by 9x+8y=180

I already found the critical point (10, 45/4). f(10,45/4)=1125/2, this is the max value. the answer book says there is no minimum value, but how do you know? for example f(20,0)=0, why isn't this the min value?

Question

use Lagrange Multipliers to find max & min of f(x,y)=5xy constrained by 9x+8y=180 

I already found the critical point (10, 45/4). f(10,45/4)=1125/2, this is the max value. the answer book says there is no minimum value, but how do you know? for example f(20,0)=0, why isn't this the min value?

zarkon · Answer

a simple way to see why it has no min is to look at the following 

$$9x+8y=180\Rightarrow8y=180-9x$$

then

$$f(x,y)=5xy=\frac{5}{8}x8y=\frac{5}{8}x(180-9x)$$

which is a parabola which opens down.  therefore it has no min.

irishboy123 · Answer

the multiplier simplifies the algebra but in my experience is pretty bad at ID'ing the nature of the extrema

turning it here into a single var problem makes that poss.  

a lot of the time, it turns on the physical nature of the function