# Real life geometry problem

#### nickorette

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
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.

#### Failure

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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