I'm faced with the following problem, find the maximum and minimum value of the function:
$$f(x,y)= x^{5}y^{7}e^{-2 x-2 y}$$
within the triangle with the corners (0,0), (3 ,0) and (0,3 ).
My initial thought is that the function cant be negative, therefore the point (0,0) must be the minimum since it gives the function value of 0.
To find the maximum I'll use the partial derivative.
$$f(x,y)= x^{5}y^{7}e^{-2 x-2 y}$$
I have solved the partial derivatives with respect to x and y. I get the point x= 5/2 and y= 7/2. Which is a point outside of the boundary. I will not include the calculations here since they were pretty tedious.
Now I'm investigating the boundary lines. My approach is to create 3 lines and investigate them for max/min values.
$$L_1 -> x = 0$$
$$L_2 -> y = 0$$
$$L_3 -> y = - x +3$$
If we look at $$L_1 -> x = 0$$ we see that y can only vary between 0 and 3.
So then we analyze $$f(0,a) = 0^{5}a^{7}e^{-2 (0)-2 a}$$
When a is between 0 and 3. We see that we will always get 0.
If we look at $$L_2 -> y = 0$$ we see that x can only vary between 0 and 3.
So then we analyze $$f(a,0) = a^{5}0^{7}e^{-2 a-2 0}$$
When a is between 0 and 3. We see that we will always get 0.
So this leaves us with the last line $$L_3 -> y = - x +3$$
$$f(x,-x+3)= x^{5}(-x+3)^{7}e^{-2 x-2 (-x+3)}$$
I guess i have to take the derivative of this? Im not sure that it is the right step and i dont know how to take the derivative of it. Can anyone point me in the right direction of what i should be doing here?
Please tell me if something was unclear.
/John
After substituting you have: $$f(x)=x^5(3-x)^7e^{-6}, \\ f'(x)=5x^4(3-x)^7e^{-6}-7x^5(3-x)^6e^{-6}=0 \Rightarrow \\ x^4(3-x)^6(5(3-x)-7x)=0 \Rightarrow \\ (x,y)=(0,3),(3,0),\left(\frac54,\frac74\right).$$