Arectangle 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?

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

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation: