how many square feet will cover a rectangle 60 feet by 20 feet? Explain how you determined your answer
1 answer:
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Answer:
Width of the archway at its base: 11 units, height of the archway at its highest point: 30.25 units.
Step-by-step explanation:
The graph that represents the equation is now sketched with the help of a graphing tool and whose representation is included in a file attached below. The point associated with x-intercepts are (-3, 0) and (8, 0) and the point associated with the highest point is (2.5, 30.25). The width of the archway at its base (
) and the height of the archway at its highest point (
) are, respectively:
Width of the archway at its base: 11 units, height of the archway at its highest point: 30.25 units.
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 =