A tent-shaped shelter has a ridged roof of length 3 metres and a horizontal rectangular floor of length x metres. Each end of the shelter is an isosceles triangle of slant side length 2 metres and base length 1 metre. The shelter, shown in the diagram below, is symmetric about vertical planes through the centre of the rectangular floor and parallel to its sides. The value of x is between 0 and 3
+ sqrt 15. (The shelter cannot exist for other values of x.)
The total volume
V (x)m3 of the shelter is given by
V
(x)= 1 /12 (2x +3) √ 15
−(x −3)2 (0 ≤x ≤3+ 15).
(You are
not asked to derive this formula.)
For parts (a) and (b) (and for part (c), if you use Mathcad there) you
should provide a printout annotated with enough explanation to make it
clear what you have done.
NB:Ifyou define
x to be a range variable in part (a) and wish to use x in
a symbolic calculation in part (b), then you will need to insert the
definition
x := x between the two parts in your worksheet.
(a) Use Mathcad to obtain the graph of the function
V (x).
(b) This part of the question requires the use of Mathcad in each sub-part.
(i) By using the differentiation facility and the symbolic keyword
‘simplify’, find an expression for the derivative
V (x).
(ii) By either applying a solve block or solving symbolically, find a
value of
x for which V (x)= 0.
(iii) Verify, by the Second Derivative Test, that this value of
x
corresponds to a local maximum of
V (x). (It should be apparent
from the graph obtained in part (a) that this is also an overall maximum within the domain of
V (x).)
(c) Using Mathcad, or otherwise, find the maximum possible volume of
the shelter, according to the model.