3cm and 4 cm if 2 sides of the rectangle lie along the legs?
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put the right angle at the origin, with the short leg along the y-axis and the long leg along the x-axis
(make sure to sketch this out, by the way)
the hypotenuse will have the equation y = (-3/4)x + 3
one vertex of the rectangle will be at the point (x , y) on the hypotenuse, which can be rewritten as (x , -3/4 x + 3)
the area of the rectangle will be xy
A = xy = x(-3/4 x + 3) = x(3 - 3/4 x) = 3x - 3/4 x^2
(subbing in the expression for y from the line)
*****
A ' = 3 - 3/2 x
A " = -3/2 , confirming the local extreme will be a max
set A ' = 0 = 3 - 3/2 x
3 = 3/2 x
x = 2
and if x = 2, then y = -3/4 (2) + 3 = -3/2 + 3 = 3/2
the rectangle will be 2 x 3/2, and have an area of 3
(note that this is half the area of the original triangle)
the key to problems like this is to place the figure on the coordinate plane so you can write equations for the unknown side
****
or, without calculus:
A = x(-3/4 x + 3)
this is a parabola that opens downward with zeros at x = 0 and x = 4
the vertex (a maximum) will be halfway between these at x = 2
and when x = 2, then y = 3/2 ==> A = 2(3/2) = 3 wow, looks familiar!
(make sure to sketch this out, by the way)
the hypotenuse will have the equation y = (-3/4)x + 3
one vertex of the rectangle will be at the point (x , y) on the hypotenuse, which can be rewritten as (x , -3/4 x + 3)
the area of the rectangle will be xy
A = xy = x(-3/4 x + 3) = x(3 - 3/4 x) = 3x - 3/4 x^2
(subbing in the expression for y from the line)
*****
A ' = 3 - 3/2 x
A " = -3/2 , confirming the local extreme will be a max
set A ' = 0 = 3 - 3/2 x
3 = 3/2 x
x = 2
and if x = 2, then y = -3/4 (2) + 3 = -3/2 + 3 = 3/2
the rectangle will be 2 x 3/2, and have an area of 3
(note that this is half the area of the original triangle)
the key to problems like this is to place the figure on the coordinate plane so you can write equations for the unknown side
****
or, without calculus:
A = x(-3/4 x + 3)
this is a parabola that opens downward with zeros at x = 0 and x = 4
the vertex (a maximum) will be halfway between these at x = 2
and when x = 2, then y = 3/2 ==> A = 2(3/2) = 3 wow, looks familiar!
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Draw a right triangle in first quadrant of coordinate plane, so that leg of length 4 is along x-axis, and leg of length 3 is along y-axis. The 3 points defining this triangle are (0,0), (4,0) and (0,3)
Hypotenuse has equation: 3x + 4y = 12 -----> y = −3/4 x + 3
Bottom left corner of rectangle is at point (0,0)
Top right corner of rectangle is on hypotenuse.
Therefore, top right corner has coordinates (x, −3/4 x + 3), for x between 0 and 4
Area of rectangle = x * (−3/4 x + 3) = −3/4 x² + 3x
Now we just need to maximize A, where A = −3/4 x² + 3x
Since this is a parabola that opens down, maximum value is located at vertex
x-coordinate of vertex = −b/(2a) = −3/(−3/2) = 2
Maximum area : A = −3/4 (2)² + 3(2) = −3 + 6 = 3 cm²
Hypotenuse has equation: 3x + 4y = 12 -----> y = −3/4 x + 3
Bottom left corner of rectangle is at point (0,0)
Top right corner of rectangle is on hypotenuse.
Therefore, top right corner has coordinates (x, −3/4 x + 3), for x between 0 and 4
Area of rectangle = x * (−3/4 x + 3) = −3/4 x² + 3x
Now we just need to maximize A, where A = −3/4 x² + 3x
Since this is a parabola that opens down, maximum value is located at vertex
x-coordinate of vertex = −b/(2a) = −3/(−3/2) = 2
Maximum area : A = −3/4 (2)² + 3(2) = −3 + 6 = 3 cm²