Real life geometry problem

Jul 2010
1
0
What are the dimensions of the largest (volume) rectangular prisim that can fit through the opening and lay flat inside of the rectangular prism pictured?



The opening is a half circle with a radius of 14.5". My free hand 3d circles aren't the greatest (Rofl)
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
I see this problem in 2 steps.

Firstly a 2D problem looking down onto the top of the box. From here you a maximising a rectangle inside a semi circle.

You need to find the area of this rectangle as a function and solve it when its derivative is equal to zero.

If the rectangle in question has dimensions y and x then \(\displaystyle A(x,y) = 2xy\) and \(\displaystyle r^2 = x^2+y^2\) then \(\displaystyle A(x) = 2x\times \sqrt{14.5^2-x^2}\) now find \(\displaystyle x\) where \(\displaystyle A'(x)=0\)

After you have this we can find the third deminsion.
 
Jul 2009
555
298
Zürich
I see this problem in 2 steps.

Firstly a 2D problem looking down onto the top of the box. From here you a maximising a rectangle inside a semi circle.

You need to find the area of this rectangle as a function and solve it when its derivative is equal to zero.

If the rectangle in question has dimensions y and x then \(\displaystyle A(x,y) = 2xy\) and \(\displaystyle r^2 = x^2+y^2\) then \(\displaystyle A(x) = 2x\times \sqrt{14.5^2-x^2}\) now find \(\displaystyle x\) where \(\displaystyle A'(x)=0\)

After you have this we can find the third deminsion.
But you are allowed to tilt the smaller prism when inserting it: so no, I don't think a reduction to a 2D problem does the job. At the very least there is a danger that by assuming that the prism will not be tilted when inserting it we will miss the optimal solution to this problem
 
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