<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>1</em><em>2</em><em>5</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>
Answer:
3
Step-by-step explanation:
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:
C=84
Step-by-step explanation:
Since you use 3 for pi. You plug the radius into the formula. C=2×3x14=84