A rectangle has its base on the x axis and its upper two vertices on the parabola y = 9 − x2. (a) Draw a graph of this problem.
(b) Label the upper right vertex of the rectangle (x,y) and indicate the lengths of the sides of the rectangle. (c) Express the area A as a function of x and state the domain of A. (d) Calculate the derviative of A and solve for the critical values. (e) What is the largest area the rectangle can have?
1 answer:
Answer:
- (a) see attached
- (b) width: 2x; height: y = 9-x²
- (c) A=2x(9-x²) . . . 0 ≤ x ≤ 3
- (d) dA/dx = -6x² +18; x=±√3
- (e) 12√3 units²
Step-by-step explanation:
(a) The attachment shows the graph of the parabola in blue. It also shows an inscribed rectangle in black.
(b) The upper right point of the rectangle is shown in the attachment as (x, y). The dimension y is the height of the rectangle. The x-dimension is half the width of the rectangle, which is symmetrical about the y-axis. Hence the width is 2x.
(c) As with any rectangle, the area is the product of length and width:
... A = (2x)(9 -x²) . . . . . the attachment shows a graph of this
... A = -2x³ +18x . . . . . expanded form suitable for differentiation
A suitable domain for A is where both x and A are non-negative: 0 ≤ x ≤ 3.
(d) The derivative of A with respect to x is ...
... A' = -6x² +18
This is defined everywhere, so the critical values will be where A' = 0.
... 0 = -6x² +18
... 3 = x² . . . . . . . divide by -6, add 3
... √3 = x . . . . . . . -√3 is also a solution, but is not in the domain of A
(e) The rectangle will have its largest area where x=√3. That area is ...
... A = 2x(9 -x²) = 2√3(9 -(√3)²) = 2√3(6)
... A = 12√3 . . . . square units . . . . ≈ 20.785 units²
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