Question 795585: I seem to be really stuck on this question.
A rectangular picture is to be surrounded by a frame. The dimension of the frame is 24 inches by 32 inches. Let w be the width of the frame.
Write an equation for the area of the picture in terms of the width of the frame
A(w)=
I've tried 4(w+24)4(w+32), 2(w+24) 2(w+36), (4w+24) (4w+36) and (2w+24)(2w+36)
The second part to the question is: If the area of the picture itself is 560 square inches, determine the width of the frame. It wants the width in inches.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! A rectangular picture is to be surrounded by a frame.
The dimension of the frame is 24 inches by 32 inches.
Let w be the width of the frame.
Write an equation for the area of the picture in terms of the width of the frame
:
A frame with a width w, subtracts 2w from the overall dimension to get the picture dimensions, therefore
A(w)= (24-2w)(32-2w)
FOIL
A(w) = 768 - 48w - 64w + 4w^2
A quadratic equation
A(w) = 4w^2 - 112w + 768
:
If the area of the picture itself is 560 square inches, determine the width of the frame.
A(w) = 560, therefore:
4w^2 - 112w + 768 = 560
4w^2 - 112w + 768 - 50 = 0
4w^2 - 112w + 208 = 0
simplify, divide by 4
w^2 - 28w + 52 = 0
Factors to
(w-26)(w-2) = 0
Two solutions
w = 26 in, not possible, throw this solution out
and
w = 2 inches is the width of the frame, makes sense
:
:
Check it by finding the area of the picture with w=2
(24-4)*(32-4) = 560, area of the picture
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