MHB Derivatives with Quadratic Functions.

[Tori No Otoko]

Slightly confused at what it wants me to do here?

 

[skeeter]

just an algebra drill ...

$T'(z) = 0 \implies \dfrac{z}{\sqrt{80^2+z^2}} = \dfrac{D-z}{\sqrt{50^2+(z-D)^2}}$

$z\sqrt{50^2+(z-D)^2} = (D-z)\sqrt{80^2+z^2}$

$z^2[50^2+(z-D)^2] = (D-z)^2[80^2+z^2]$

$50^2z^2 + z^2(z-D)^2 = 80^2(D-z)^2 + z^2(D-z)^2$

$0 = 80^2(D-z)^2 - 50^2z^2$

$0 = 80^2(D^2 - 2zD + z^2) - 50^2z^2$

$0 = 80^2D^2 - 80^2 \cdot 2zD + (80^2-50^2)z^2$

$0 = 6400(D^2-2zD) + 3900z^2$

$0 = 64(D^2-2zD) + 39z^2$

$0 = 39z^2 - 128zD + 64D^2$

 

[HOI]

Notice that the problem gives you T', not T. It is not necessary to differentiate. Just set it equal to 0 and do the algebra to get the final result.