Find the minimum value of the expression radical(a^2+4)+radical((3-a)^2+16)

Answer 1

Take the 1st derivative and set to 0: Use chain rule and linearity property of derivatives.

(1/2)(a²+4)^-1/2(2a) + (1/2)((3-a)²+16)^-1/22(3-a)(-1) = 0 = a/(a²+4)^1/2 - (3-a)/((3-a)²+16)^1/2

Separate and square:

a²/(a²+4) - (3-a)²/((3-a)²+16)

multiply to remove fractions (essentially cross-multiplying)

a²((3-a)²+16) = (3-a)²(a²+4)

a²(3-a²) + 16a² = (3-a)²a²+(3-a)²4

Phew! We see some cancelling terms...

16a² = (3-a)²4 (Leaving a factorable quadratic which I'll leave to you.)

Try the two critical points to find the absolute min. Note that +/- infinity the function goes to infinity, so there are no absolute maxes.

Please consider a tutor. Take care.