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3 years ago

Solve each system by elimination. (Help! I’ll mark you brainly)

Mathematics

1 answer:

zheka24 [161]3 years ago

4 0

Answer:

2) y=-1 4)y=-1

Step-by-step explanation

4x+8y=20

+ -4x+2y=-30

---------------------

10y=-10

isolate y by dividing 10 on each side

y=-1

-6x+5y=1

+ 6x+4y=-10

----------------------

9y=-9

y=-1

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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

Charra [1.4K]

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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