We're minimizing

subject to

. Using Lagrange multipliers, we have the Lagrangian

with partial derivatives

Set each partial derivative equal to 0:

Subtracting the second equation from the first, we find

Similarly, we can determine that

and

by taking any two of the first three equations. So if

determines a critical point, then

So the smallest value for the sum of squares is

when

.
Answer:
You need to attach something to this?
Step-by-step explanation:
I could help you but theres nothing here to see?
Answer:
Option C is correct, i.e. 2x² +7x -4.
Step-by-step explanation:
Given is (2x-1)(x+4)
Applying FOIL method:-
F- multiply First two, O- multiply Outer two, I- multiply Inner two, L- multiply Last two.
So, (2x-1)(x+4) = 2x*x +2x*4 -1*x -1*4
(2x-1)(x+4) = 2x² +8x -1x -4
(2x-1)(x+4) = 2x² +7x -4
Hence, option C is correct, i.e. (2x-1)(x+4) = 2x² +7x -4.
Answer:
387
Step-by-step explanation: