Answer:
The width of the frame is 0.0746 meters
Step-by-step explanation:
We are told that after the frame has been attached to the solar collector, the area that is left exposed is [tex]2.5m^2[/tex] . As we see in the figure, the dimensions of this area are
[tex](w-2x)[/tex] and [tex](l-2x)[/tex]
where [tex]x[/tex] is the width of the frame.
The product of these dimensions must equal the exposed area:
[tex](w-2x)(l-2x)=2.5m^2[/tex]
Now since [tex]w=1.5m[/tex] and [tex]l=2m[/tex] we have:
[tex](1.5-2x)(2-2x)=2.5[/tex]
we expand this and solve for x using the quadratic formula:
[tex]4x^2-7x+3=2.5[/tex]
[tex]4x^2-7x+0.5=0[/tex]
we get two solutions:
[tex]x=1.6753[/tex]
[tex]x=0.0746[/tex]
We take the second solution i.e [tex]x=0.0746[/tex], because first one gives a width larger than the dimensions of the solar collector which cannot be possible.
Thus the width of the frame is 0.0746 meters.
