1. ## Analysis

100m of angle steel is used to make a rectangular frame with three crossbars in the figure.
Figure attached

a) if the width of the frame is xm, determine an expression for l, the length of the frame in metres, in terms of x.

b) the frame is to be covered by light aluminium sheeting. if the area of this sheeting is Am^2, formulate a rule of A in terms of x.

c) Sketch the graph of A against x stating the axes intercepts and turning point. you will need to consider an appropriate domain.

d) Find the maximum area and the value of x which gives this area.

2. Originally Posted by scubasteve94
100m of angle steel is used to make a rectangular frame with three crossbars in the figure.
Figure attached

a) if the width of the frame is xm, determine an expression for l, the length of the frame in metres, in terms of x.

b) the frame is to be covered by light aluminium sheeting. if the area of this sheeting is Am^2, formulate a rule of A in terms of x.

c) Sketch the graph of A against x stating the axes intercepts and turning point. you will need to consider an appropriate domain.

d) Find the maximum area and the value of x which gives this area.
1. From your sketch you know:

$\displaystyle 100 = 2l + 4x$ Solve for l. (50-2x)

2. The area of a rectangle is calculated by: $\displaystyle area = length \cdot width$

Plug in the terms for length and width and simplify. ($\displaystyle a(x)=-2x^2+50x$)

3. The graph of a is a parabola opening down. Thus the maximum value is at it's vertex. Calculate the coordinates of the vertex and you'll get the x-value which correspond with the maximum of a. $\displaystyle \left( \frac{25}2\ ,\ \frac{625}2\right)$