Imagine you want to find the maximum area a rectangle can take when it is bounded on one of its edges by the x-axis and its opposite vertices touch a parabola (y = - x^2 + 6a^2 , where (-root6)*a < x < (root6)*a. I managed to solve this using derivatives and found the maximum area (in terms of a). I also plotted the graphs for various values of 'a', and determined the maximum area this way (though this way is not very rigourous, as I can only check for a few values of 'a'. Does anyone know of another method to do this problem that doesnt require the use of calculus?
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calculus needed for this problem? (1 Viewer)
- Thread starter Loner
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jb_nc
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Have you looked at the graph? I used a program to graph it up and it looks as though it may me possible to find the area under the parabola using the formula of an ellipse.Loner said:
Root of 6 is where the parabola cuts the x-axis making it a nice and clean function if you're using the formula.
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Yes I checked it, and plotted for various values of 'a'. It's definately a parabola. From plotting several graphs I can calculate the maximum area of the rectangle in terms of 'a'. Though this is not very rigorous as you can't calculate for ALL values of a. Could one then prove by indiction that it's true for all values of 'a'?
I can also do it using differentiation calculating an expression for the area of the rectnagle and then maximising it. Another way I did it was calculate the area under the parabola using integration, subtracting the area of the rectangle from this, then doing differentiation to minimise the area.
I thought perhaps there is another way using geometry, as the grapical method can't test for all vaues of 'a'.
Any other takers? Thanks..
I can also do it using differentiation calculating an expression for the area of the rectnagle and then maximising it. Another way I did it was calculate the area under the parabola using integration, subtracting the area of the rectangle from this, then doing differentiation to minimise the area.
I thought perhaps there is another way using geometry, as the grapical method can't test for all vaues of 'a'.
Any other takers? Thanks..