Difficult Optimisation problem (maximizing a cuboid)

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The discussion focuses on optimizing the volume of a cuboid defined by a rectangle with a variable width and height derived from the equation of a circle. The height is expressed as 1296 - x², where x ranges from 0 to 36. Participants explore the conditions under which the height is at least 75% of the width, leading to the formulation of the area as a function of x. The use of calculus is suggested to find the optimum dimensions for maximizing the area of the cuboid.

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  • Understanding of basic calculus, specifically derivatives
  • Familiarity with the equation of a circle and its geometric implications
  • Knowledge of optimization techniques in mathematical contexts
  • Ability to manipulate algebraic expressions and inequalities
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  • Learn about geometric properties of ellipses and their equations
  • Explore the concept of area maximization in calculus
  • Investigate inequalities and their applications in geometric contexts
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Difficult Optimisation problem! (maximizing a cuboid)

Find derivate d(x)
 
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Well, just cut in in half on the y-axis. The base of half your cuboid (... a rectangle or square in this case...) is just x. The height is 1296-x^2. We're looking on x=0 to 36, right? Well, almost. At what value of x is the height going to be at least 75% of the base? Then what's the area of this half rectangle? Can you find the optimum using calculus?
 


Ok, it made it easier for me to think about but forget about what i said about cutting it in half.

If x were greater than 36, then you would be outside the ellipse, but you want to say inside.

Think of creating your rectangle using the variable x. If I set one corner at (x,0), then I can set another corner at (x, sqrt(1296-x^2)). So it has width 2x and height sqrt(1296-x^2). Therefore the area is ____ ?

Regarding the 75% thing - for what x will the height be exactly 75% of the width? Now for what values is it less than 75% of the width? Think that you want sqrt(1296-x^2) to be less than 75% of 2x.
 

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