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Maximising and minimising

Maximising and minimising

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Maximising and minimising

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  1. Free-Standing Mathematics Activity Maximising and minimising

  2. How can they design containers to: • hold as much as possible • use as little material as possible? Manufacturers use containers of different shapes and sizes In this activity you will use calculus to solve such problems

  3. Maximum volume of a box Step 1 – find a formula A piece of card measures 20 cm by 20 cm A square with sides of length x cm is removed from each corner An open-topped box is made by folding the card Think about What are the dimensions of the box? Can you find an expression for the volume of the box?

  4. Maximum volume of a box Volume of the box V = x(20 – 2x)(20 – 2x) V = x(400 – 80x + 4x2) V = 400x – 80x2 + 4x3 20 – 2x 20 Think about What must be done next to find the maximum value of V? How can you be sure this gives the maximum? What else could you do?

  5. Maximum volume of a box Step 2 – use calculus Find the first and second derivatives Solve the equation = 0 This gives the values of x for which the function V has turning points Then work out the maximum value of V Use the second derivative to check that this is a maximum (and not a minimum)

  6. Maximum volume of a box – step 2 V = 400x – 80x2 + 4x3 Let V = V(x), then the first derivative= V ′(x) and the second derivative = V ″(x) V ′(x) = 400 – 160x + 12x2 V ″(x) = -160 + 24x V has turning points where V ′(x) = 0 So 400 – 160x + 12x2 = 0

  7. To simplify 400 – 160x + 12x2 = 0, divide by 4 100 – 40x + 3x2 = 0 Maximum volume of a box Think about Would you solve this using factors or the formula? The determinant, b2 – 4ac = (-40)2 – 4 × 3 × 100 = 1600 – 1200 = 400 = 202 A square number means there are factors: 100 – 40x + 3x2 = (10 – x)(10 – 3x) = 0 Think about Why not? So x = 10 or But x = 10 is not a reasonable solution So x = is the only one that can apply

  8. Maximum volume of a box Doesx = give a maximum? V″(x) = -160 + 24x When x = V″ ( ) = -160 + 24 × = -80 < 0 So this does give a maximum Vmax = 400( ) – 80( )2 + 4( )3 Vmax = 593 cm3 (3 sf)

  9. What if you wanted the minimum material to make a cylinder with a required volume? In this case you would have two variables (radius and height) and one fixed quantity (volume) Think about Why is having two variables a problem? In order to differentiate, you need an expression for the quantity you want to minimise (or maximise) in terms of just one variable

  10. First, use the fixed volume to eliminate one of the variables (either the height or radius) When you have an expression for the quantity of material needed to make the cylinder in terms of just one variable, differentiate it and put the derivative = 0 Solve this equation to find the value of the variable that gives a minimum (or maximum) Then find the value of the other variable and the minimum (or maximum) that you require Working with a cylinder

  11. Say you want to find the minimum metal needed to make a can to hold 500 ml (the same as 500 cm3) If r cm is the radius and h cm is the height,the volume V = πr2h and the metal used M = 2πr2 + 2πrh Minimum material to make a can Think about Why is this the area of metal needed? So if V = 500 then πr2h = 500 It is easier to eliminate h(because r is a squared term) Think about Which variable is it easier to get rid of? h =

  12. Minimum material to make a can Using h = M = 2πr2 + 2πrh = 2πr2 + 2πr× 250 π M = 2πr2 + 2π × = 2πr2 + 1000r-1 So = = 4πr − 1000r-2 500 When = 0 there will be turning point dM dM πr dr dr So 4πr − 1000r-2 = 0 giving 4πr = 1000r-2 4πr3 = 1000 Think about Why is r= -4.30 not included here? r3 = r = 4.30 (3 sf)

  13. r = 4.30 (3 sf) Does this give a minimum? M ′(r) = 4πr − 1000r-2so M ″(r) = 4π + 2000r-3 When r = 4.30, M ″(r) is positive Minimum material to make a can So this will give a minimum value for M Now h = = 500 π (π×4.302) = 8.60 (3 sf) So M = 2πr2 + 2πrh = 2π×4.302 + 2π× 4.30 × 8.60 = 349 cm2 (3 sf) Think about If r were eliminated instead of h, would this answer be the same?

  14. Now set your own problem Solve a packaging problem that needs: • either to use the least materials • or to hold the most volume

  15. You could adapt one of the examples With a can, you might decide that the material used in the base and top needs to be double thickness, so you would end up with a different answer or you might decide that two different metals should be used with a different unit cost for each

  16. Boxes come in all sorts of shapes, with and without lids − what about a Toblerone box? Or how about swimming pools, with a shallow end and a deep end? Would the cement be equally thick all over? What about ice cream cone packaging? What about a wooden play house? (Remember the door!) There are many other possibilities http://mathforum.org/dr.math/faq/formulas/ gives lots of other formulae

  17. Summary of method If you have only one variable and one given value, Think about What do you do? use these to create a formula for the term that is to be minimised/maximised Think about What do you do? If you have two variables and one given value, first decide how to write one of the variables in terms of the other variable and the given value Only then create the formula for minimisation/maximisation

  18. Summary of method Now you have an expression for the quantity you want to maximise/minimise Think about What do you do next? Differentiate it with respect to the variable, put the result equal to 0 and solve the equation you get To ensure you have a valid answer to the problem, find the second derivative of the expression Substitute each value of the variable to see which, if any, gives the required maximum/minimum Think about What is the rule that tells you which you have? Finally work out the dimensions you need and then the quantity you wanted to maximise/minimise