Question

A freight company has shipping orders for two products. The first product has a unit volume of 10 cu ft, and it weighs 50 lbs. The second product's unit volume is 3 cu ft, and it weighs 40 lbs. If the company's trucks have 2300 cu ft of space and can carry 21,000 lbs., how many units of each can be transported in a single shipment with one truck?

Answer 1

Answer:

116 units of the first product

380 units of the second product

Step-by-step explanation:

Product 1 has a unit volume of  10 cu ft

Product 2 has a unit volume of 3 cu ft

The truck has 2300 cu ft of space

Product 1 weighs 50 lbs

Product 2 weighs 40 lbs

The truck can carry 21000 lbs

Let X be the units of product 1

Let Y be the units of product 2

The given information can be expressed as:

10X+3Y=2300...(1)

50X+40Y=21000...(2)

Solving the system of equations:

10X+3Y=2300...(1)

10X=2300-3Y

X=(2300-3Y)/10

Substituting X in (2) we have:

50X+40Y=21000

50[(2300-3Y)/10]+40Y=21000

50[(2300/10)-(3Y/10)]+40Y=21000

50[230-(3Y/10)+40Y=21000

11500-(150Y/10)+40Y=21000

11500-15Y+40Y=21000

11500+25Y=21000

25Y=21000-11500

25Y=9500

Y=380

Substituting Y in (1) we have:

10X+3Y=2300...(1)

10X+3(380)=2300

10X+1140=2300

10X=2300-1140

10X=1160

X=116

So 116 units of the first product and 380 units of the second product can be transported in a single shipment with one truck.

Answer 2

The number of units of the first product that can be transported is 116, and the number of units of the second product that can be transported is 380.

Given to us,

• the volume of the first product = 10 ft³
• weight of the first product = 50 lbs.
• the volume of the second product = 3 ft³
• weight of the second product = 40 lbs.

• the volume truck can carry = 2,300 ft³
• weight truck can carry = 21,000 lbs.

Assumption

Let's assume the number of units of the first product that can fit in the truck is x, and the number of units of the second product that can fit in the truck is y.

Total volume truck can carry

Total volume truck can carry

= (Total volume of the first product for x number of units) +(Total volume of second product for y number of units)

$2,300 = 10x+3y\\\\x =\dfrac{2300-3y}{10}$

Total weight truck can carry

Total weight truck can carry

= (Total weight of the first product for x number of units) +(Total weight of the second product for y number of units)

$21000 = 50x + 40y$

substituting the value of x,

$21000 = 50(\dfrac{2300-3y}{10}) + 40y$

$21000 = 5\times({2300-3y}) + 40y$

$21000 = 11500-15y + 40y$

$21000 -11500= -15y + 40y$

$y=\dfrac{9500}{25}=380$

substituting the value of y in x's equation,

$x =\dfrac{2300-3y}{10}$

$2,300 = 10x+3y\\\\x =\dfrac{2300-(3\times 380)}{10}$

$x = 116$

Hence, the number of units of the first product that can be transported is 116, and the number of units of the second product that can be transported is 380.

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