Originally Posted by

**JeffM** I am not sure that this helps because it uses calculus.

$d > 0,\ e > 0\, f > 0, \text { and } d + e + f = 28.$

$z = def - y(28 - d - e - f) \implies$

$\dfrac{\delta z}{\delta d} = ef - y \implies ef = y \text { if } \dfrac{\delta z}{\delta d} = 0.$

$\dfrac{\delta z}{\delta e} = df - y \implies df = y \text { if } \dfrac{\delta z}{\delta e} = 0.$

$\dfrac{\delta z}{\delta f} = de - y \implies de = y \text { if } \dfrac{\delta z}{\delta f} = 0.$

$\dfrac{\delta z}{\delta y} = 28 - d - e - f \implies d + e + f = 28 \text { if } \dfrac{\delta z}{\delta y} = 0.$

$ef = y \implies y > 0 \text { and } f = \dfrac{y}{e}.$

$df = y \implies d * \dfrac{y}{e} = y \implies \dfrac{d}{e} = 1 \implies d = e.$

$de = y \text { and } f = \dfrac{y}{e} \implies f = \dfrac{de}{e} = d = e \implies$

$d = e = f = \dfrac{28}{3}.$

In other words, to maximize (d * e * f) subject to (d + e + f) = 28, d = e = f = 28/3. How you discover that with algebra, I am not sure, but I suppose you can confirm it with algebra.

$\therefore a = \dfrac{28}{3} < 9.5,\ b = \sqrt{\dfrac{28}{3}} < 3.1 , \text { and } c = \sqrt[4]{\dfrac{28}{3}} < 1.8.$

$\therefore abc < 9.5 * 3.1 * 1.8 = 53.01 \le 64.$