An-open-box-of-area-486cm-2-If-the-length-is-twice-the-breadth-Find-the-maximum-volume-of-the-box-hence-Show-the-volume-is-maximum-

Question Number 14543 by chux last updated on 02/Jun/17

An open box of area {486 cm}^{2} . If the

length is twice the breadth . Find

the maximum volume of the box .

hence, Show the volume is maximum .

Commented by chux last updated on 02/Jun/17

please help

Answered by sandy_suhendra last updated on 22/Jun/17

let the breadth = x

the length = 2 x

and the height = h

the area of the box = 2 x . x + (2 x + x + 2 x + x) h

486 = {2 x}^{2} + 6 hx

6 hx = 486 - {2 x}^{2}

h = \frac{486 - {2 x}^{2}}{6 x} = \frac{243 - x^{2}}{3 x}

the volume = V = 2 x . x . h = {2 x}^{2} (\frac{243 - x^{2}}{3 x}) = 162 x - \frac{2}{3} x^{3}

\frac{dV}{dx} = 162 - {2 x}^{2} = 0 \Rightarrow x = 9

V \max = 162 \times 9 - \frac{2}{3} \times 9^{3} = 972 {cm}^{3}

Answered by Tinkutara last updated on 02/Jun/17

l = 2 b

Let height = h

Since box is open, area will exclude

the topmost surface area . Hence,

2 h (l + b) + l b = 486

6 b h + 2 b^{2} = 486

b^{2} + 3 b h = 243

3 b h = 243 - b^{2}

h = \frac{243 - b^{2}}{3 b}

V = l b h = 2 b^{2} h = 2 b (\frac{243 - b^{2}}{3})

= \frac{486 b - 2 b^{3}}{3}

\frac{d V}{d b} = \frac{1}{3} (486 - 6 b^{2}) = 0 gives b = 9

\frac{d^{2} V}{{d b}^{2}} = \frac{1}{3} (- 12 b) which is always

negative since b > 0 .

Volume is maximum at b = 9 cm

V_{\max} = \frac{486 \times 9 - 2 \times 729}{3}

= 486 \times 3 - 2 \times 243 = 972 {cm}^{3}