Question
A doorway has the shape of a parabolic arch and is 16 feet high at the center and 8 feet wide at the base. If a rectangular box 12 feet high must fit through the doorway, what is the maximum width the box can have?
Answers
First find the rule of the parabola:
We know that the parabola curves (is open) towards the bottom, so the rule is
y(x)=b-ax²
if we set the y-axis to cut the vertex.
Since
y(0)=16=b, so b=16
or y(x)=16-kx²
Also,
y(4)=0 tells us that k=1, or
the rule of the doorway is:
y(x)=16-x²
At
y(x)=8, we have
8=16-x²
Solve for x to get x=√8=2√2.
Thus the box will touch the doorway at
x=±2√2,
The maximum box size is double 2√2, or
4√2.
We know that the parabola curves (is open) towards the bottom, so the rule is
y(x)=b-ax²
if we set the y-axis to cut the vertex.
Since
y(0)=16=b, so b=16
or y(x)=16-kx²
Also,
y(4)=0 tells us that k=1, or
the rule of the doorway is:
y(x)=16-x²
At
y(x)=8, we have
8=16-x²
Solve for x to get x=√8=2√2.
Thus the box will touch the doorway at
x=±2√2,
The maximum box size is double 2√2, or
4√2.
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