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Vsevolod [243]
2 years ago
8

Kira is trying to drink more water and juice each day. The difference in the amount of water in a jug and the amount of juice in

the bottle she is drinking from is 192 ounces. She has consumed a total of 42 ounces, which is StartFraction 3 Over 32 EndFraction of the bottle of juice and StartFraction 9 Over 64 EndFraction of the jug of water. Which system of equations can be used to determine the total number of ounces in the jug of water, x, and the total number of ounces in the bottle of juice, y? x minus y = 192 and StartFraction 9 Over 64 EndFraction x StartFraction 3 Over 32 EndFraction y = 42 x minus y = 192 and StartFraction 3 Over 32 EndFraction x StartFraction 9 Over 64 EndFraction y = 42 x y = 192 and StartFraction 9 Over 64 EndFraction x StartFraction 3 Over 32 EndFraction y = 42 x y = 192 and StartFraction 3 Over 32 EndFraction x StartFraction 9 Over 64 EndFraction y = 42.
Mathematics
1 answer:
luda_lava [24]2 years ago
8 0
That just started to confuse me
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by
11111nata11111 [884]

Answer:

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is V=l\times b\times h

V=(11-2x)\times (7-2x)\times x

V=4x^3-36x^2+77x

Derivate w.r.t x,

V'(x)=4(3x^2)-2(36x)+77

V'(x)=12x^2-72x+77

The critical point when V'(x)=0

12x^2-72x+77=0

Solve by quadratic formula,

x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}

x=4.607,1.392

Derivate again w.r.t x,

V''(x)=24x-72

Now, V''(4.607)=24(4.607)-72=38.568>0 (+ve)

V''(1.392)=24(1.392)-72=-38.592 (-ve)

So, there is maximum at x=1.392.

The length of the box is l=11-2x

l=11-2(1.392)=8.216

The breadth of the box is b=7-2x

b=7-2(1.392)=4.216

The height of the box is h=1.392.

The dimension of the open rectangular box is 8.216\times 4.216\times 1.392.

The volume of the box is V=l\times b\times h

V=8.216\times 4.216\times 1.392

V=48.217\ in.^3

The volume of the box is 8.217 cubic inches.

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