Answer:
Verified
Step-by-step explanation:
Let A matrix be in the form of
Then det(A) = ad - bc
Matrix A transposed would be in the form of:
Where we can also calculate its determinant:
det(AT) = ad - bc = det(A)
So the determinant of the nxn matrix is the same as its transposed version for 2x2 matrices
Answer:
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
