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harkovskaia [24]
4 years ago
12

You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3

cents/cm2 and the top and bottom cost 4 cents/cm2. Find the dimensions of the box that minimize the total cost of the materials used.
Mathematics
1 answer:
Charra [1.4K]4 years ago
6 0

Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

h=\frac{504}{lw}

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides C_{s}= 3[2(l + w)h] = 6(l + w)h

C_{s}= 3[2(l + w)h]

C_{s}=6(l+w)(\frac{504}{lw} )

Cost of the top and the bottom C_{(t,p)}= 4(2lw) = 8lw

Total cost of the box C = 3024\frac{(l+w)}{lw} + 8lw

                                      = 3024[\frac{1}{l}+\frac{1}{w}] + 8lw

To minimize the cost of the sides

\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0

\frac{3024}{l^{2}}=8w

\frac{378}{l^{2}}=w ---------(1)

\frac{dC}{dw}=3024(-w^{-2})+8l=0

\frac{3024}{w^{2}}=8l

\frac{378}{w^{2}}=l

w^{2}=\frac{378}{l}-------(2)

Now place the value of w from equation (1) to equation (2)

(\frac{378}{l^{2}})^{2}=\frac{378}{l}

\frac{(378)^{2} }{l^{4}}=\frac{378}{l}

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

w^{2}=\frac{378}{7.23}

w^{2}=52.28

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

h=\frac{504}{(7.23)^{2}}

h = 9.64 cm                        

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