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dusya [7]
3 years ago
10

A 120-inch strip of metal 12 inches wide is to be made into a small open trough by bending up two sides on the long side, at rig

ht angles to the base. The sides will be the same height, x. If the trough is to have a maximum volume, how many inches should be turned up on each side?
Mathematics
1 answer:
mezya [45]3 years ago
3 0

Answer:

x= 3 inch should be turned up on each side

Step-by-step explanation:

Let the height of trough be x.

Width of trough be 12 - 2x.

and length of trough = 120 inch

Volume of trough, V = L×W×H = 120 × (12-2x) × x = 120x(12 - 2x)

For maximum volume, we find V' = 0

i.e 1440 -480x = 0

or  x = \frac{1440}{480}

or  x= 3

Hence x= 3 inch should be turned up on each side

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Can you help me with my homework
goldfiish [28.3K]

Answer:

4. Option C (3a+2) inches

5. Extraneous solution x=-5/6 because we get for width and length negative values.

Solution: Value of x is 3

Length of the box: 5 ft

Width of the box: 4 ft

Step-by-step explanation:

4. Area of a rectangle: A=12a^2-a-6 square inches

Width: w=4a-3

Length: l=?

A=w l

Replacing A by 12a^2-a-6 and w by 4a-3

12a^2-a-6 = (4a-3) l

Solving for l: Dividing both sides of the equation by 4a-3:

(12a^2-a-6) / (4a-3) = (4a-3) l / (4a-3)

Simplifying:

(12a^2-a-6) / (4a-3) = l

l = (12a^2-a-6) / (4a-3)

Factoring the numerator:

12a^2-a-6 = (4a-3)(3a+2)

Let's check it:

(4a-3)(3a+2)=4a(3a)+4a(2)-3(3a)-3(2)=12a^2+8a-9a-6→(4a-3)(3a+2)=12a^2-a-6

Replacing the numerator:

l = (4a-3)(3a+2) / (4a-3)

Simplifying:

l = (3a+2) inches


5. Length: l=(3x-5) ft

Width: w=(2x-1) ft

Height: h=2 ft

Volumen of the box: V=40 ft^3

x=?

Length: l=?

Width: w=?

V = l w h

Replacing the given:

40 ft^3 = (3x-5) ft (2x-1) ft 2 ft

40 ft^3 = 2 (3x-5)(2x-1) ft^3

40=2(3x-5)(2x-1)

Dividing both sides of the equation by 2:

40/2=2(3x-5)(2x-1)/2

Simplifying:

20=(3x-5)(2x-1)

Eliminating the parentheses on the right side of the equation applying the distributive property:

20=3x(2x)+3x(-1)-5(2x)-5(-1)

20=6x^2-3x-10x+5

Adding like terms:

20=6x^2-13x+5

Equaling to zero: Subtracting 20 from both sides of the equation:

20-20=6x^2-13x+5-20

0=6x^2-13x-15

6x^2-13x-15=0

ax^2+bx+c=0; a=6, b=-13, c=-15

Using the quadratic formula:

x=[-b+-sqrt(b^2-4ac)]/(2a)

x=[-(-13)+-sqrt((-13)^2-4(6)(-15))]/(2(6))

x=[13+-sqrt(169+360)]/12

x=[13+-sqrt(529)]/12

x=[13+-23]/12

x1=(13-23)/12=(-10)/12=-10/12=-(10/2)/(12/2)→x1=-5/6

x2=(13+23)/12=36/12→x2=3


With x=-5/6

l=(3x-5) ft

l=(3(-5/6)-5) ft

l=(-5/2-5) ft

l=-(5/2+5) ft

l=-(5+2(5))/2 ft

l=-(5+10)/2 ft

l=-15/2 ft < 0. The length cannot be a negative number then x=-5/6 is a extraneous solution.

w=(2x-1) ft

w=(2(-5/6)-1) ft

w=(-5/3-1) ft

w=-(5/3+1) ft

w=-(5+3(1))/3 ft

w=-(5+3)/3 ft

w=-8/3 ft <0. The width cannot be a negative number then x=-5/6 is a extraneous solution.


With x=3

l=(3x-5) ft

l=(3(3)-5) ft

l=(9-5) ft

l=4 ft

w=(2x-1) ft

w=(2(3)-1) ft

w=(6-1) ft

w=5 ft

and h=2 ft

Let's check the volume

V= w l h

V=(5 ft)(4 ft)(2 ft)

V=40 ft^3 Correct

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