Author Topic: Help for possible SAC question (Read 575 times) Tweet Share

missemtee · « **on:** May 15, 2013, 08:52:33 pm »

Tried this but getting the wrong answer! (Maximum and Minimum problems when the functions unknown)
The frame of a container in the shape of a cuboid (x=width, 2x=height, L=length) is to be made with a total length of 18 metres of steel edging. Find:
a) The value of L in terms of x
b) The expression of the volume in terms of x only
c) The length of each edge for maximum volume
d) The maximum volume
THANKS

Daenerys Targaryen · « **Reply #1 on:** May 15, 2013, 09:35:28 pm »

(a) Let P=perimeter
P=4x+4(2x)+4L
P=18
$12x+4L=18$
$L=\frac{1}{4}(18-12x)$

(b) V=x(2x)L
But $L=\frac{1}{4}(18-12x)$
$V=2x^2(\frac{1}{4}(18-12x))$
$V=\frac{x^2}{2}(18-12x)$
$V=9x^2-6x^3$

(c) $\text{for max V, }\frac{dV}{dx}=0$
$\frac{dV}{dx}=18x-18x^2$
$18x-18x^2=0$
$x(1-x)=0$
$x=0$ and $x=1$ but $x>0$ because length
$\therefore width=1m \text{ height}=2m \text{ length}=\frac{3}{2}m$

(d) $V=w \times l \times h$
$V=1 \times 2 \times \frac{3}{2}$
$V=3m^3$

EDIT: fixed the error awks

Jeggz · « **Reply #2 on:** May 15, 2013, 09:56:45 pm »

Quote from: HatersGonnaHate on May 15, 2013, 09:35:28 pm

(a) Let P=perimeter
P=4x+4(2x)+4L
P=18
$10x+4L=18$
$L=\frac{1}{4}(18-10x)$

(b) V=x(2x)L
But $L=\frac{1}{4}(18-10x)$
$V=2x^2(\frac{1}{4}(18-10x))$
$V=\frac{x^2}{2}(18-10x)$
$V=9x^2-5x^3$

(c) $\text{for max V, }\frac{dV}{dx}=0$
$\frac{dV}{dx}=18x-15x^2$
$18x-15x^2=0$
$3x(6-5x)=0$
$x=0$ and $x=\frac{6}{5}$
$\therefore width=\frac{6}{5}m \text{ height}=\frac{12}{5}m \text{ length}=\frac{3}{2}m$

(d) $V=w \times l \times h$
$V=\frac{6}{5} \times \frac{12}{5} \times \frac{3}{2}$
$V=4.32m^3$

In part a) I think you meant $12x+4L=18$ instead of $10x$

Daenerys Targaryen · « **Reply #3 on:** May 15, 2013, 09:57:50 pm »

Aq indeed ill edit it (damn now everything after is wrong) poo! -.-!

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