SECTION B (PART 1) A curve is given by y = 8x + \(\frac{27}{2x^2}\), FIND: (a) an expression for \(\frac{dy}{dx}\), (b) the coordinates of the stationary po...

Given:  y = 8x + \(\frac{27}{2x^2}\) =  y = 8x + \(\frac{27}{2}\)x\(^{-2}\),

(a) \(\frac{dy}{dx}\) = 8x\(^0\) +  \(\frac{27}{2}\)(-2)x\(^{-3}\)

= 8 - 27x\(^{-3}\) = 8 -  \(\frac{27}{x^3}\)

(b) \(\frac{dy}{dx}\) = 8 -  \(\frac{27}{x^3}\) = 0

= 8 = \(\frac{27}{x^3}\)

= 8x\(^3\) = 27

x\(^3\) = \(\frac{27}{8}\)

x = \( \sqrt[3]{\frac{27}{8}}\) = \(\frac{3}{2}\)

x = \(\frac{3}{2}\)

y = 8x + \(\frac{27}{2x^2}\) = 8(\(\frac{3}{2}\)) + \(\frac{27}{2}\)(\(\frac{3}{2}\))\(^{-2}\)

y = 4(3) +  \(\frac{27}{2}\)(\(\frac{4}{9}\))

y = 12 + 6 = 18

Thus, stationary points are (\(\frac{3}{2}\), 18)

\(\frac{d^2y}{dx^2}\) = - \(\frac{27}{x^3}\) = - 27(-3)x\(^{-4}\) = \(\frac{81}{x^4}\) = \(\frac{81}{(\frac{3}{2})^4}\) = 81 x \(\frac{16}{81}\) = 16 > 0

For all x, the stationary point is a minimum.

(c) gradient = 8 -  \(\frac{27}{x^3}\) = 8 -  \(\frac{27}{2^3}\) = 8 -  \(\frac{27}{8}\) = \(\frac{37}{8}\)

Equation of normal, at (2, 2)

but m\(_2\) =  \(\frac{- 1}{m_1}\) = \(\frac{- 1}{\frac{37}{8}}\) = \(\frac{- 8}{37}\)

y - y\(_1\) = m\(_2\)[x - x\(_1\)]

y - 2 = \(\frac{- 8}{37}\)[x - 2]

37y - 74 = - 8x + 16 

37y + 8x - 90 = 0