SOLUTION: A norman window has the shape of a rectangle surmounted by a semicircle of a diameter equal to the width of the rectangle. If the permimeter of the window is 20 feet, what dimensio
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-> SOLUTION: A norman window has the shape of a rectangle surmounted by a semicircle of a diameter equal to the width of the rectangle. If the permimeter of the window is 20 feet, what dimensio
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Question 403095: A norman window has the shape of a rectangle surmounted by a semicircle of a diameter equal to the width of the rectangle. If the permimeter of the window is 20 feet, what dimensions will admit the most light (maximize the area)? Answer by solver91311(24713) (Show Source):
Let represent the measure of the vertical dimension of the rectangular portion of the window. Let represent the horizontal dimension of the window. is also the diameter of the semi-circular part of the window.
The perimeter of the window is then:
But we are given that the perimeter is 20 feet, so:
Solve for :
The area of the window is:
Substitute to create a function for area in terms of
Simplify:
Two ways to go from here:
Algebra Method
This is a quadratic with a negative lead coefficient, hence the graph is a parabola that opens downward. That means the vertex is a maximum. Recall that the vertex of is at .
For this problem:
and a little arithmetic gets us to:
And the radius of the semi-circle is then:
Calculus Method
This is a 2nd degree polynomial equation, hence continuous and twice differentiable over the real numbers. Therefore there will be a local extremum at any point where the first derivative is equal to zero and that point will be a maximum if the 2nd derivative is negative at that point.
Hence
is the -coordinate of the maximum point.
The dimension and the semi-circle radius are calculated as above.
John
My calculator said it, I believe it, that settles it
