Question

y varies jointly as a and b and inversely as the square root of c y=96 when a=8 b=6 c=16 find y when a=7 b=5 c=25

Answer 1

y = 56

The initial statement is y varies as
`(ab)/(√(c) )`

To convert to an equation multiply by k the constant of variation

y =
`(kab)/(√(c) )`

y = 96 when a = 8 , b = 6 , c = 16

y =
`(kab)/(√(c) )`

hence k =
`(y√(c) )/(ab)`

=
`(96√(16) )/(8 x 6)` =
`(96 x 4)/(8 x 6)` = 8

Thus y =
`(8ab)/(√(c) )`

given a = 7 , b = 5 , c = 25 then

y =
`(8 x 7 x 5)/(√(25) )` = 8 × 7 = 56

Answer 2

`\bf \qquad \qquad \textit{compound proportional variation} \\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{\underline{y} varies jointly as \underline{a} and \underline{b} and inversely as }√(c)}{y=\cfrac{kab}{√(c)}}`

`\bf \textit{we also know that } \begin{cases} y=96\\ a=8\\ b=6\\ c=16 \end{cases}\implies 96=\cfrac{k(8)(6)}{√(16)}\implies 96=\cfrac{48k}{4} \\\\\\ \cfrac{96\cdot 4}{48}\implies 8=k\qquad therefore\qquad \boxed{y=\cfrac{8ab}{√(c)}} \\\\\\ \textit{when } \begin{cases} a=7\\ b=5\\ c=25 \end{cases}\textit{ what is \underline{y}?}\implies y=\cfrac{8(7)(5)}{√(25)}\implies y=\cfrac{8(7)(5)}{5}\implies y=56`