Area of only the picture = (20-2x)(10-2x)
(can you see why I subtracted 2x ?)
of course both of these dimensions have to be positive, and of course x has to be positive,
so 10-2x > 0
-2x > -10
x < 5
so 0 < x < 5 , can you see the max/mins in that?
A framed picture has a total length and breadth of 20 cm and 10 cm. The frame has width x cm.
a) Find the rule for the area (A cm^2) of the picture inside.
b) What are the minimum and maximum values of x?
4 answers
Yes! Thank you for your help :)
But what happened to the (20-2x)?
Answering "But what happened to the (20-2x)?"
It would be x < 10. HOWEVER, if x < 10 is substituted to (10-2x), then it would not make sense anymore and it would result in a negative answer.
E.g. sub x = 9
(20-2(9))(10-2(9)) = 2*-8 = -16
So this would not be the answer as measurement or the length CANNOT be a negative. It has to be positive.
x < 10 is not the correct answer.
It would be x < 10. HOWEVER, if x < 10 is substituted to (10-2x), then it would not make sense anymore and it would result in a negative answer.
E.g. sub x = 9
(20-2(9))(10-2(9)) = 2*-8 = -16
So this would not be the answer as measurement or the length CANNOT be a negative. It has to be positive.
x < 10 is not the correct answer.