Should i use dy\dx (normal differentiation) or d²y\dx² (differentiation of the differentiation)?

Use both!

You want to minimize P, so differentiate P with respect to x and set the derivative equal to 0 and solve for any critical points.

P = 8/x + 2x

dP/dx = -8/x² + 2 = 0

8/x² = 2

x² = 8/2 = 4

x = ± √4 = ± 2

You can then use the second derivative to determine the concavity of P, and its sign at a given critical point decides whether it is a minimum or a maximum.

We have

d²P/dx² = 16/x³

When x = -2, the second derivative is negative, which means there's a relative maximum here.

When x = 2, the second derivative is positive, which means there's a relative minimum here.

So, P has a relative maximum value of 8/(-2) + 2(-2) = -8 when x = -2.

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

happy to

pls mark as brainliest

13664

step-by-step explanation:

=cross multiply

=100 x = 13664