Solve the problem.
-$\text { Find the maximum value of } f ( x , y , z , w ) = x + y + z + w \text { subject to } x ^ { 2 } + y ^ { 2 } + z ^ { 2 } + w ^ { 2 } = 1 \text {. }$
A) 10
B) 4
C) 2
D) 1

Question

Quizplus · Accepted Answer

We can use the Cauchy-Schwarz inequality to solve this problem.
Applying Cauchy-Schwarz, we have:
$$(1+1+1+1)(x^2+y^2+z^2+w^2) \geq (x+y+z+w)^2$$
Simplifying gives:
$$4 \geq (x+y+z+w)^2$$
Taking the square root of both sides, we get:
$$2 \geq x+y+z+w$$
So the maximum value of $f(x,y,z,w)=x+y+z+w$ is $\boxed{2}$, achieved when $x=y=z=w=\frac{\sqrt{4}}{4}$.

Solve the Problem A) 10 B) 4 C) 2 D) 1

Solve the Problem A) 10
B) 4
C) 2
D) 1