I have an investigation and I am stuck on the question:
Investigate further rectangular bases with different length:width ratios.
Form a conjecture about the effect of this length:width ratio on the value of x needed to maximise the volume.
My conjectures for the other two parts images are attached.
Part iii The last section of this part is to find a conjecture based on a rectangular cake tin of width w cm, which when a square is cut from each corner of length x cm, a maximum volume for the resulting open top cake tin will be obtained. This is done by using the methods of expansion, differentiation, and the quadratic formula. 1 = 21 : V(x) = (w -2x) (1 -2x)x x Where w = 5 1 - 2x :. V(x) = (5-2x) (1 -2x)x W W - 2x V(x) = x(5 - Lx -21x + 4x2) V(x) = -1x2 - 21x2 + 4x3 V(x) =7-31x2 +4x3 Maths Methods Stage one SACE ID: 913106W :. V'(x) = 7-61x + 12x2 V' (x) = 12x2 - 61x + 512 = 0 The derivative of V'(x) can be substituted into the quadratic formula x = =bvb-4ac 2a " where a, b and c are numerical coefficients of the quadratic equation. Calculations Sign diagram Findings 6l + V3612 - 2412 .: the local minimum can be found by using x = * = Max point 24 1(3- V3) 61 + V1212 12 and the local maximum can be found when X = = 24 12 x = 13+v3). This equation represents all values of 61 + 21V3 I and x up to infinity. There are many number X= 24 possibilities, factors and variables that can affect 31+ 1V3 he value of x. As well as infinite combinations of x = _ 12 ratios between w and I. The conjecture made 31+ 1V3 above will be investigated for further evidence. X = 12 0 31 + 1V3 - OR X = 31 - 1V3 1(3+V3) x= 12 12 12 (3+ V3) (3 - V3) 12 - OR x = 12 Using the conjecture to discover the value of x When w = 7 When w = 106.54 When w = - the conjecture produced the same 1(3+ V3) _7(3 + v3) 1(3 + V3) 106.54(3 + V3) 15 results as part ii. 12 12 12 12 1(3 + 13) 15 ( 3 + V3 ) 12 12Part iii The last section of this part is to find a conjecture based on a square cake tin with a side length of I cm, which when a square is cut from each corner of length x cm, a maximum volume for the resulting open top cake tin will be obtained. This done by using the methods of expansion, differentiation, and the quadratic formula. V (x) = (1- 2x) X (1 - 2x) X (x) V(x) = x(12 - 4x1+ 4x2) V(x) = x/2 - 4x21 + 4x3 V'(x) = 12x2 - 8lx - 12 The derivative of V'(x) can be substituted into the quadratic formula x = -b+vbz-4ac where a, b and c are numerical coefficients of the quadratic equation. 2a Calculations Sign diagram Findings 81+ V(-81)2 - 4(12) (1)2 .: the local minimum can be found by using x = V ( x ) = Max point 2(12) and the local maximum can be found when x = 81 + V6412 - 4812 V (x) = - This equation represents all values of I and x up to 24 infinity. There are many number possibilities, z19LA F 18 V (x) =- factors and variables that can affect the value of x. 24 The conjecture made above will be investigated for 81 + 41 V(x) = further evidence. 24 21+ 1 V(x) = - 6 0 OR X=; Using the conjecture to discover the value of x When I = 1 When I = 2 When I = 3 When I = IT When I = 4 the conjecture produced the same results as part ii. 1 1 1 2 3 1 4 al - 6 6 6 6 1 6 6 6 6 x = 0.16 x = 0.33 x = 0.52 X = X = 0.66
