Optimisation Problems
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Question 1 of 5
1. Question
An open rectangular box is to be made by cutting out square corners from a square `90xx90`cm piece of cardboard and folding up the sides. What is the maximum volume of the box? `V=` (54000)`\text(cm)^3`
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Volume of a Rectangular Prism
`V=lxxwxxh`First, check the current dimensions of the boxLength (`l`)`=902x`Width (`w`)`=902x`Height (`h`)`=x`Form a function by substituting the current dimensions into the Volume Formula`V` `=` `lxxwxxh` Volume Formula `=` `(902x)(902x)x` Substitute values `=` `(8100180x180x+4x^2)x` `=` `(8100360x+4x^2)x` `V` `=` `4x^3360x^2+8100x` Arrange the terms in descending powers Start optimizing the function by getting its first derivative and solving for `x``V` `=` `4x^3360x^2+8100x` `V’` `=` `12x^2720x+8100` `=` `12(x^260x+675)` `=` `12(x45)(x15)` `=` `0` `x=``45,15` Substitute each value of `x` to the second derivative of `V``V’` `=` `12x^2720x+8100` `V”` `=` `24x720` `=` `24(``45``)720` `=` `1080720` `=` `360` This value is positive, which means `45` yields the minimum value`V’` `=` `12x^2720x+8100` `V”` `=` `24x720` `=` `24(``15``)720` `=` `360720` `=` `360` This value is negative, which means `15` yields the maximum valueFinally, compute for the maximum volume by substituting `x=15` to `V``V` `=` `4x^3360x^2+8100x` `=` `4(15^3)360(15^2)+8100(15)` Substitute `x=15` `=` `4(3375)360(225)+8100(15)` `=` `1350081000+121500` `=` `54 000\text(cm)^3` `54 000\text(cm)^3` 
Question 2 of 5
2. Question
A `24`cm piece of wire is bent into the shape of a rectangle. Find the maximum area of this rectangle. `A=` (36)`\text(cm)^2`
Hint
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Perimeter of a Rectangle
`P=2l+2w`Area of a Rectangle
`A=lw`First, draw a diagram to represent the problemLength (`l`)`=x`Width (`w`)`=y`Substitute the current dimensions into the Perimeter FormulaRecall that the wire is `24`cm long, hence `P=24``P` `=` `2l+2w` Perimeter Formula `24` `=` `2x+2y` Substitute values `24``:2` `=` `(2x+2y)``:2` Divide both sides by `2` `12` `=` `x+y` `12x` `=` `y` `y` `=` `12x` Leave only `y` on the left side Now, form a function by substituting `y=12x` to the Area Formula`A` `=` `lw` Area Formula `A` `=` `xy` Substitute `x` and `y` `A` `=` `x(12x)` Substitute `y=12x` `A` `=` `12xx^2` Start optimizing the function by getting its first derivative and equating it to `0` to solve for `x``A` `=` `12xx^2` `A’` `=` `122x` `=` `0` `12` `=` `2x` `12``:2` `=` `2x``:2` Divide both sides by `2` `6` `=` `x` `x` `=` `6` Substitute this value of `x` to the second derivative of `A``A’` `=` `122x` `A”` `=` `2` This value is negative, which means `x=6` yields the maximum valueFinally, compute for the maximum area by substituting `x=6` to `A``A` `=` `12xx^2` `=` `12(6)(6)^2` Substitute `x=6` `=` `7236` `=` `36\text(cm)^2` `36\text(cm)^2` 
Question 3 of 5
3. Question
Maximize the cube below given that the sum of its dimensions is `60\text(cm)` (8000)`\text(cm)^3`
Hint
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Volume of a Cube
`V=lwh`Form the first equation using the given information`x+x+y` `=` `60` Sum of dimensions equal `60\text(cm)` `2x+y` `=` `60` Equation `1` Form another equation using the Volume Formula`V` `=` `lwh` Volume Formula `V` `=` `x*x*y` Substitute given dimensions `V` `=` `x^2y` Equation `2` Since we are hoping to minimize the Volume, choose Equation `2` as the main functionWrite Equation `1` in terms of `y` and use it so the main equation only has one variable`2x+y` `=` `60` Equation `1` `y` `=` `602x` Substitute this to Equation `2``V` `=` `x^2y` Equation `2` `V` `=` `x^2(602x)` Substitute `y` from previous step `V` `=` `60x^22x^3` Start optimizing the function by getting its first derivative and equating it to `0` to solve for `x``V` `=` `60x^22x^3` `V’` `=` `120x6x^2` `120x6x^2` `=` `0` Equate `V’` to `0` `6x(20x)` `=` `0` Factor out `6x` `x` `=` `0,20` Since `x` is a length, it cannot be `0`. Hence, we choose `x=20`Substitute this value of `x` to the second derivative of `V``V’` `=` `120x6x^2` `V”` `=` `12012x` `=` `12012(``20``)` Substitute `x=20` `=` `120240` `=` `120` This value is negative, which means `x=20` yields the maximum valueCompute for `y` by substituting `x` to Equation `1``2x+y` `=` `60` Equation `1` `2(20)+y` `=` `60` Substitute `x=20` `40+y` `=` `60` `y` `=` `6040` `y` `=` `20` Finally, compute for the maximum volume by substituting `x` and `y` to `V``x=20``y=20``V` `=` `x^2y` `=` `(20)^2(20)` Substitute `x` and `y` `=` `20^3` `=` `8000\text(cm)^3` `8000\text(cm)^3` 
Question 4 of 5
4. Question
A company that manufactures soft drinks wants to save money by restricting each can to have a Surface Area of `726pi\text(cm)^2`. Find the maximum volume of the can with this restriction.Hint
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Volume of a Cylinder
`V=pir^2h`Surface Area of a Cylinder
`SA=2pir^2+2pirh`Form the first equation using the given informationWrite this equation in terms of `h``SA` `=` `2pir^2+2pirh` Surface Area Formula `726pi` `=` `2pir^2+2pirh` Substitute given value `726pi2pir^2` `=` `2pirh` `2pirh` `=` `726pi2pir^2` `2pirh``:2pi` `=` `(726pi2pir^2)``:2pi` Divide both sides by `2pi` `rh` `=` `363r^2` `rh``:r` `=` `(363r^2)``:r` Divide both sides by `r` `h` `=` `(363r^2)/r` Form another equation using the Volume Formula and substituting `h``V` `=` `pir^2h` Volume Formula `=` `pir^2((363r^2)/r)` Substitute `h` from previous step `=` `pir(363r^2)` `V` `=` `363pirpir^3` Start optimizing the function by getting its first derivative and equating it to `0` to solve for `x``V` `=` `363pirpir^3` `V’` `=` `363pi3pir^2` `363pi3pir^2` `=` `0` Equate `V’` to `0` `363pi` `=` `3pir^2` `363pi``:pi` `=` `3pir^2``:pi` Divide both sides by `pi` `363` `=` `3r^2` `363``:3` `=` `3r^2``:3` Divide both sides by `3` `121` `=` `r^2` `sqrt121` `=` `sqrt(r^2)` Get the square root of both sides `11` `=` `r` `r` `=` `11` Substitute this value to the second derivative of `V``V’` `=` `363pi3pir^2` `V”` `=` `6pir` `=` `6pi(11)` Substitute `r=11` `=` `66pi` This value is negative, which means `r=11` yields the maximum valueFinally, compute for the maximum volume by substituting `r` to `V``V` `=` `363pirpir^3` `=` `363pi(11)pi(11^3)` Substitute `r` `=` `3993pi1331pi` `=` `2662pi\text(cm)^3` `2662pi\text(cm)^3` 
Question 5 of 5
5. Question
A page is printed with a margin, as shown below. Given that the page area is `288\text(cm)^2`, maximise the dimensions so that the printed internal area is maximised.
`x=` (12)`y=` (24)
Hint
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Perimeter of a Rectangle
`P=2x+2y`Area of a Rectangle
`A=xy`Form the first equation using the Area Formula for the Page Area`A` `=` `xy` Area Formula `288` `=` `xy` `xy` `=` `288` Equation `1` Form another equation using the Area Formula for the Printed Area`A` `=` `xy` Area Formula `A_p` `=` `(x11)(y22)` Substitute given dimensions `A_p` `=` `(x2)(y4)` Equation `2` Since we are hoping to maximize the Printed Area, choose Equation `2` as the main functionWrite Equation `1` in terms of `y` and use it so the main equation only has one variable`xy` `=` `288` Equation `1` `xy``:x` `=` `288``:x` Divide both sides by `x` `y` `=` `288/x` Updated Equation `1` Substitute this to Equation `2``A_p` `=` `(x2)(y4)` Equation `2` `=` `(x2)(288/x4)` Substitute `y` from previous step `=` `x(288/x)4x2(288/x)+8` `=` `2884x576/x+8` `=` `296576/x4x` `A_p` `=` `296576x^(1)4x` Updated Equation `2` Start optimizing the function by getting its first derivative and equating it to `0` to solve for `x``A_p` `=` `296576x^(1)4x` `A’_p` `=` `(576x^(2))4` `A’_p` `=` `576/(x^2)4` `576/(x^2)4` `=` `0` Equate `A’_p` to `0` `576/(x^2)` `=` `4` `576/(x^2)``times x^2` `=` `4``times x^2` Multiply both sides by `x^2` `576` `=` `4x^2` `576``:4` `=` `4x^2``:4` Divide both sides by `4` `144` `=` `x^2` `sqrt144` `=` `sqrt(x^2)` Get the square root of both sides `12` `=` `x` `x` `=` `12` Substitute this value of `x` to the second derivative of `A_p``A’_p` `=` `576x^(2)4` `A”_p` `=` `1152x^(3)` `=` `1152/(x^3)` `=` $$\frac{1152}{\color{#007DDC}{12}^3}$$ Substitute `x=12` `=` `1152/1728` This value is negative, which means `x=12` yields the maximum valueCompute for `y` by substituting `x` to the Updated Equation `1``y` `=` `288/x` Updated Equation `1` `y` `=` `288/12` Substitute `x=12` `y` `=` `24` `x=12``y=24` 