Application of calculus to find the minimum cost
A frame of a shelter is made of rods of two different lengths:
- $x$ metres for top and bottom edges
- $y$ metres for each sloping edge
The frame is to be covered by a rectangular sheet of material.
Total area of sheet is $24m^2$.
- Show the total length, $L$ metres of the rods used in a shelter is given by: $L = 3x + {\frac{48}x}$
- To minimise costs, the total length of rods used in a frame must be as small as possible . Calculate the minimum cost of a frame.
For part 1, I would say that
Area = $2xy = 24$
$y = \frac{12}x$
The frame is made up of $3x$ rods for top and bottom and $4y$ for the side rods.
$L = 3x + 4(\frac{12}x)$
$L = 3x + \frac{48}x$
For part 2, stationary points occur when $\frac{dy}{dx} = 0$
So I should take the derivative of $f(x) = 3x + \frac{48}x$
Which is $3 - 48x^{-2}$
I'm not sure if this is right or how to proceed from here.
$\endgroup$ 11 Answer
$\begingroup$You just need to set the derivative equal to zero now.
$$3 - \frac{48}{x^2} = 0$$ $$\therefore \frac{3x^2 - 48}{x^2} = 0$$ Assuming that $x \neq 0$, $$ 3x^2 - 48 = 0 $$ $$\therefore x^2 = 16$$ $$\therefore x = 4$$
and subsequently, $$ y = \frac{12}{x} $$ $$ \therefore y = 3 $$
Hence the minimum total length of frame is, $$ L_{min} = 3x + \frac{48}{x} $$ $$ L_{min} = 24 $$
Hence the minimum cost of the frame is now, $$Cost_{min} = L_{min}*(cost \hspace{4pt} per \hspace{4pt} metre)$$ Or based on the area the cost can be calculated as, $$ Cost_{min} = 24 * (cost \hspace{4pt} per \hspace{4pt} metre^2)$$
$\endgroup$ 3
