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?
But what happened to the (20-2x)?
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?
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.
Yes! Thank you for your help :)
a) The area (A) of the picture inside the frame can be calculated by subtracting the area of the frame from the total area of the frame and picture:
A = (20 - 2x) * (10 - 2x) square cm
b) To find the minimum and maximum values of x, we need to consider the limits of the frame's width. Since the frame cannot have negative dimensions, we can set up the following inequalities:
0 ≤ 20 - 2x ≤ 10
0 ≤ 20 - 2x ≤ 20
Simplifying these inequalities, we get:
0 ≤ -2x ≤ -10
0 ≤ -2x ≤ 0
Dividing all sides of the inequalities by -2, we have:
0 ≤ x ≤ 5
0 ≤ x ≤ 0
Now, we need to discard the second inequality since it is trivial (0 ≤ 0). Therefore, the valid range for x is:
0 ≤ x ≤ 5
Thus, the minimum value for x is 0 cm (no frame), and the maximum value for x is 5 cm (frame covering half of each side).
a) To find the area of the picture inside the frame, we need to subtract the area of the frame from the total area of the frame and the picture.
The total area of the frame and the picture can be found by multiplying the length and breadth:
Total area = length * breadth
Total area = 20 cm * 10 cm
To find the area of the frame, we need to subtract the area of the picture from the total area:
Area of the frame = Total area - Area of the picture
Since the frame has a uniform width of x cm on all sides, we can calculate the length and breadth of the picture by subtracting twice the width of the frame from the original length and breadth.
Length of the picture = Length - 2 * Width
Length of the picture = 20 cm - 2 * x
Breadth of the picture = Breadth - 2 * Width
Breadth of the picture = 10 cm - 2 * x
Therefore, the area of the picture can be calculated as:
Area of the picture = Length of the picture * Breadth of the picture
Substituting the expressions for length and breadth of the picture, we get:
Area of the picture = (20 cm - 2 * x) * (10 cm - 2 * x)
So the rule for the area of the picture inside the frame is:
A = (20 - 2x)(10 - 2x)
b) To find the minimum and maximum values of x, we need to consider the dimensions of the frame and the picture.
Since x represents the width of the frame, it cannot be negative. Therefore, x ≥ 0.
To find the minimum value of x, we need to consider the case when the picture inside the frame has the maximum possible area. This occurs when the width of the picture is minimized, which means the width of the frame is maximized. In this case, x will be the maximum value that still allows the total width of the frame and the picture to be equal to 20 cm.
If we set the width of the picture to 0 (meaning no picture inside the frame), then we can find the maximum possible width of the frame:
Total width = Width of the picture + Width of the frame
Total width = 0 + Width of the frame
Total width = Width of the frame
Since the total width is given as 20 cm, we have:
Width of the frame = Total width = 20 cm
Therefore, the maximum value of x is 20 cm.
The minimum value of x is 0, as the width of the frame cannot be negative.
So the minimum value for x is 0 cm and the maximum value for x is 20 cm.