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Fish Tank Optimization

Fish Tank Optimization. Megan Satawa. Problem. When I head to college this fall, I want to take a fish-tank to have African Clawed Frogs in my dorm because they’re an awesome aquatic pet!. Problem.

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Fish Tank Optimization

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  1. Fish Tank Optimization Megan Satawa

  2. Problem When I head to college this fall, I want to take a fish-tank to have African Clawed Frogs in my dorm because they’re an awesome aquatic pet!

  3. Problem The filter I have is for a 10-gallon tank. The cost of the tank increases as the surface area increases. Therefore, to minimize cost, I want to find the dimensions I need to have a 10-gallon tank of the minimum surface area possible.

  4. Variables Two edges of the base will have a length of x, and the opposite two a length of 2x. The height will be y. y x 2x

  5. Volume Since the volume of a rectangular prism is length x width x height, I can use a volume equation of: V = 2x²y

  6. Volume V = 2x²y This equation can be put in into terms of only x to become y = V/2x²

  7. Volume y = V/2x² The volume of the tank must be 10 gallons, which is equal to 2,310 in³ (1 gal. = 231 in³.) This can be plugged into the equation for V. y = 2310/2x²

  8. Surface Area Next, an equation for surface area is needed. This can be obtained by plugging the variables of the tank’s dimensions into surface area formulas for each face of the box (l x w.) SA = 2x² + 4xy + 2xy

  9. Surface Area & Volume SA = 2x² + 4xy + 2xy Then, the equation for y found earlier can be plugged into the surface area equation so there is only one equation to work with. SA = 2x² + 4x(2310/2x²) + 2x(2310/2x²)

  10. Simplification SA = 2x² + 4x(2310/2x²) + 2x(2310/2x²) SA = 2x² + (4620/2) + (2310/x) SA = 2x² + 6930/x) SA = 2x² + 6930x^-1

  11. Derive Then the equation can be derived and set equal to 0 to solve for x. This x- value with give a zero on the derivative graph, which means there is a critical point on the original function’s graph.

  12. Solve for X SA = 2x² + 6930x^-1 SA’ = 4x – 6930x^-2 0 = 4x – 6930x^-2 (6930/x²) = 4x 6930 = 4x³ x = 12.0104

  13. Check: Algebraically We can make sure this answer is a minimum and not a maximum by finding the value of a derivative function a little to either side of the discovered x-value. SA’(11) = -13.3 SA’(13) = 11 Since the derivative graph goes from negative to positive at the x-value, the critical point of the function graph is a minimum.

  14. Check: Graphically We can also see on the graph of the function that there is a minimum at x = 12.

  15. Solution While finding the x-value gives 2 of the 3 dimensions needed, we still need a height. This can be found by plugging the x value into the equation for y. y = 2310/2x² Y = 2310/2(12)² Y = 8.0208 in.

  16. Solution With these answers, it can be concluded that the dimensions of the 10-gallon tank of minimum surface area will be 24 x 12 x 8.0208 in. 8.0208 12 24

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