Max/min question (1 Viewer)

[enigma_1]

 

[bottleofyarn]

You may have used the wrong intercepts. When I used (0,4) instead of (0,4+a^2) as the y intercept, I got a=2. Try it, remembering that the point is in the first quadrant.

 

[ChillTime]

Your x intercept should be x1 = (a^2+4)/2a; y intercept should be y1 = a^2+4

Area:
A = (x1)(y1)/2 = (a^2+4)*(a^2+4)/2a = 0.5*a^3 + 4a + 8*a^-1

Next you differentiate the ugly expression, and set dA/dx = 0 to find your stationary point(s)

A' = 1.5*a^2 - 8*a^-2 + 4 = 0
1.5a^4 + 4a^2 -8 = 0
Let u = a^2
1.5u^2 + 4u -8 = 0
u = -4, u = 4/3
a = -2/sqrt(3) or +2/sqrt(3)

a = +2/sqrt(3)

Answer is right. 2/sqrt(3) = 2*sqrt(3)/3

2nd derivative

A" = 16a^-3 + 3a
A"(2/sqrt(3)) = 8sqrt(3) > 0, so minimum value