Simple optimization problem

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Aion

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In a factory, open plate containers (i.e. without lid) should be made which can hold [MATH]10 \: m ^3[/MATH]. The base surface should be a rectangle with twice the length than width, and the sides should be rectangular. Determine the length, width and height so that as little plate as possible can be used. You do not need to take into account spills from the material.

Is the following reasoning correct?

Total volume [MATH]V = 10 \: m ^3[/MATH]. We are given the basearea of the cubic is [MATH]2w^2 \: m ^2[/MATH] and the height [MATH]h[/MATH] is unknown.

The constraint is [MATH]2w ^2h = 10 \: m ^3[/MATH] and hence [MATH]h =5w^{-2}\: m ^3[/MATH].

Surfacearea of the cuboid [MATH]S = 2w ^2 + 6wh= 2w ^2 +30w^{-1} [/MATH] and equals the total amount of metal necessary to construct the container. By takeing derivitive of S we can find the local minimum.

[MATH]S^\prime = 4w+30w^{-2} [/MATH]
[MATH]S^\prime = 0 \iff 4w+30w^{-2} = 0 \iff w^3=7.5 \iff w = (7.5)^{1/3} \approx 1.957\: m [/MATH]
[MATH]h = 5w^{-2} = \frac{5}{7.5^{2/3}} \approx 1.305\: m[/MATH]
Therefore, to minimize the amount of metal necessary when constructing the platecontainer it should have the following dimenstions: width [MATH] w \approx 1.957 \: m[/MATH], length [MATH] \ell = 2w \approx 3.914 \: m[/MATH] and height [MATH]h \approx 1.305\: m [/MATH].
 
Aion said:
In a factory, open plate containers (i.e. without lid) should be made which can hold [MATH]10 \: m ^3[/MATH]. The base surface should be a rectangle with twice the length than width, and the sides should be rectangular. Determine the length, width and height so that as little plate as possible can be used. You do not need to take into account spills from the material.

Is the following reasoning correct?

Total volume [MATH]V = 10 \: m ^3[/MATH]. We are given the basearea of the cubic is [MATH]2w^2 \: m ^2[/MATH] and the height [MATH]h[/MATH] is unknown.

The constraint is [MATH]2w ^2h = 10 \: m ^3[/MATH] and hence [MATH]h =5w^{-2}\: m ^3[/MATH].

Surfacearea of the cuboid [MATH]S = 2w ^2 + 6wh= 2w ^2 +30w^{-1} [/MATH] and equals the total amount of metal necessary to construct the container. By takeing derivitive of S we can find the local minimum.

[MATH]S^\prime = 4w+30w^{-2} [/MATH]
[MATH]S^\prime = 0 \iff 4w+30w^{-2} = 0 \iff w^3=7.5 \iff w = (7.5)^{1/3} \approx 1.957\: m [/MATH]
[MATH]h = 5w^{-2} = \frac{5}{7.5^{2/3}} \approx 1.305\: m[/MATH]
Therefore, to minimize the amount of metal necessary when constructing the platecontainer it should have the following dimenstions: width [MATH] w \approx 1.957 \: m[/MATH], length [MATH] \ell = 2w \approx 3.914 \: m[/MATH] and height [MATH]h \approx 1.305\: m [/MATH].
Click to expand...
You have the wrong S'
If w=approx 1.957>0, How can 4w+30w^-2 =0, it must be positive!
You made two errors that cancelled out, but none the less you have two errors.
 
[MATH]S = 2w ^2 +30w^{-1} \: m ^2 [/MATH][MATH]S^\prime = 4w-30w^{-2} [/MATH][MATH] S^\prime = 0 \iff w = (7.5)^{1/3} \approx 1.957[/MATH]
Im not sure about the units. How do we know its [MATH]m [/MATH] and not [MATH]m ^2 [/MATH]?
 
Aion said:
[MATH]S = 2w ^2 +30w^{-1} \: m ^2 [/MATH][MATH]S^\prime = 4w-30w^{-2} [/MATH][MATH] S^\prime = 0 \iff w = (7.5)^{1/3} \approx 1.957[/MATH]
Im not sure about the units. How do we know its [MATH]m [/MATH] and not [MATH]m ^2 [/MATH]?
Click to expand...
What do you think the units should be for the width of this cuboid? You can also attach the units for the whole problem to see if you get the correct units in the end, otherwise you know you made a mistake.
 
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