Let B be between a and C use the segment addition postulate to solve for W
1 answer:
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Answer:
The square-based box with the greatest volume under the condition that the sum of length, width, and heigth does not exceed 174 in is a cube with each edge of 58 in and a volume of
Step-by-step explanation:
For this problem we have two constraints, that are as follows:
1) Sum of length, width, and heigth not exceeding 174 in
2) Lenght and width have the same measure (square-based box)
We know that volume is equal to the product of all three edges, and with the two conditions into account we have the next function:
The interval of interest of the objective function is [0, 87]
This problem requieres that we maximize the function that defines the volume. We start calculating the derivative of the function, wich is:
We need to remember that the derivative of a function represents the slope of said function at a given point. The maximum value of the function will have a slope equal to zero.
So we find the value in wich the derivative equals zero:
The first value () will leave us with a 'height-only box', so the answer must be
The value is between the interval of interest.
And, once we solve for the constraints, we have that:
Lenght = Width = Heigth = 58 in
Volume =
Answer:
It weighs 40 pounds on mars
Step-by-step explanation:
Direct proportion is the relationship between two variables when their ratio is equal to a constant value
- It can be expressed as y = k x, where y varies directly as x and k is constant of variation
- It can be expressed as
∵ The weight of an object on mars varies directly as the weight
of the object on earth
∴ w(m) = k × w(e)
∵ The weight of the object on earth is 75 pound
∴ w(e) = 75
∵ The weight of this object on mars is 25 pounds
∴ w(m) = 25
- Substitute them in the equation of variation
∵ 25 = k × 75
- Divide both sides by 75
∴ = k
∴ w(m) = × w(e)
∵ An object weighs 120 pounds on earth
∴ w(e) = 120
- Substitute it in the equation of variation
∴ w(m) = × 120
∴ w(m) = 40
∴ It weighs 40 pounds on mars