A company makes concrete bricks shaped like rectangular prisms. Each brick is 15 inches long, 10 inches wide, and 5 inches tall.
1 answer:
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We have a rectangle with length L that is 3 inches more than the width W. Then we can write this as:
The area of the rectangle is 180 square inches.
We have to find the width W.
As the area is equal to the product of the length and the width, we can write this equation and solve for W as:
We have a quadratic equation. The roots of this equation will be the mathematical solutions.
We can find the roots using the quadratic formula:
The solutions are W = -15 and W = 12.
The first one is not valid, as W has to be greater than 0.
Then, the solution to our problem is W = 12 in.
Answer: the width is W = 12 inches.
Answer:
a. A(x) = (1/2)x(9 -x^2)
b. x > 0 . . . or . . . 0 < x < 3 (see below)
c. A(2) = 5
d. x = √3; A(√3) = 3√3
Step-by-step explanation:
a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...
A(x) = (1/2)(x)(y)
A(x) = (1/2)(x)(9-x^2)
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b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).
On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).
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c. A(2) = (1/2)(2)(9 -2^2) = 5
The area is 5 when x=2.
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d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).
