Find Largest Area Enclosed by Rectangle w/ Given Parameters

[merikukri]

Please Help Pre-Cal !

1.A rectangle has one vertex on the line y=L-Fx,x>0 , another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area A that can be enclosed by the rectangle. Show all your work and include a sketch with labels of all important features. (Hint: Start by drawing a diagram, label the base of the rectangle as x, and write the area, A, as a function of x. …)
Where F = 5, L =4

SOL : A = f(x)= x * (-5x+4) ; the x is the base while y(x)=-5x+4 is the
height.
and the domain of x is 0<x<(4/5)
To get critical point x= 4/10 which is
where the maximum occurs at y= .8 ( took the derivative )



2.A drum in the shape of a right circular cylinder is required to have a volume of cubic centimeters. The top and bottom are made of material that costs F¢ per square centimeter; the sides are made of material that costs L¢ per square centimeters. (Hint: The formula for the volume of a right circular cylinder is V=pie r^2 , where r is the radius of the circular base and h is the height of the cylinder. The surface area of the sides can be determined by cutting the cylindrical shell vertically and flattening it out to get a rectangle whose dimensions can be determined.)


a)Express the total cost C of the material as a function of the radius r of the cylinder.
b)What is the cost if the radius is 25 cm?
c)Graph C=C(r) . Using the graph, for what value of r, approximately, is the cost C least?
Where F = 5, L =4

SOl : C= L(2*pi*r*h)+F(2*pi*r^2);
C(r) = L(2*pi*r*(2*pi*r/sqrt(3))+F(2*pi*r^2); Here height is
expressed in terms of r . The ratio used to find h in terms of r is
30-60-90 degree triangle; (2*pi*r)/(h) = sqrt(3)/1

3.Design a polynomial function with the following characteristics: degree 6; exactly four real zeros, one of multiplicity 3 at x= -F ; y-intercept at F, behaves like y = -Lx^6 for large values of |x| . Give the formula and a complete graph

Where F = 5, L =4

?


4.Create a rational function that has the following characteristics: crosses the x-axis at F; touches the x-axis at -L ; has one vertical asymptote at x=-L-1 and another at x=F+1 ; and has one horizontal asymptote, y=F . Give the formula and a complete graph

Where F = 5, L =4


SOL: Crosses X-Axis @ g
touches X-Axis @ -4
Vertical asymtote X= -4-1 = -5
Veetical asymtote x= 5+1 = 6
Horizontal asymtote y=5

5X 1
--- X ----
(X+5) (X-6)

 

[HallsofIvy]

1.A rectangle has one vertex on the line y=L-Fx,x>0 , another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area A that can be enclosed by the rectangle. Show all your work and include a sketch with labels of all important features. (Hint: Start by drawing a diagram, label the base of the rectangle as x, and write the area, A, as a function of x. …)
Where F = 5, L =4
SOL : A = f(x)= x * (-5x+4) ; the x is the base while y(x)=-5x+4 is the
height.
and the domain of x is 0<x<(4/5)
To get critical point x= 4/10 which is
where the maximum occurs at y= .8 ( took the derivative )

Yes, you done that one correctly.

2.A drum in the shape of a right circular cylinder is required to have a volume of cubic centimeters. The top and bottom are made of material that costs F¢ per square centimeter; the sides are made of material that costs L¢ per square centimeters. (Hint: The formula for the volume of a right circular cylinder is V=pie r^2 , where r is the radius of the circular base and h is the height of the cylinder. The surface area of the sides can be determined by cutting the cylindrical shell vertically and flattening it out to get a rectangle whose dimensions can be determined.)
a)Express the total cost C of the material as a function of the radius r of the cylinder.
b)What is the cost if the radius is 25 cm?
c)Graph C=C(r) . Using the graph, for what value of r, approximately, is the cost C least?
Where F = 5, L =4
SOl : C= L(2*pi*r*h)+F(2*pi*r^2);
C(r) = L(2*pi*r*(2*pi*r/sqrt(3))+F(2*pi*r^2); Here height is
expressed in terms of r . The ratio used to find h in terms of r is
30-60-90 degree triangle; (2*pi*r)/(h) = sqrt(3)/1

Were you given other information you didn't tell us? There is nothing in what you gave here that says h must be any particular function of r.
(And why do you keep saying "F= 5, L= 4"?

3.Design a polynomial function with the following characteristics: degree 6; exactly four real zeros, one of multiplicity 3 at x= -F ; y-intercept at F, behaves like y = -Lx^6 for large values of |x| . Give the formula and a complete graph
Where F = 5, L =4
?

You understand, do you not that, that there are an infinite number of correct answers for this? The problem says "Design a polynomial function". You are just asked to come up with one like this. Since the polynomial is to have a zero of degree 3 at x= -F, it must have factors (x-(-F))³= (x+F)³. It has just one more real root, let's make that x= 1, for simplicity: it has a factor of (x-1). The other two roots must be complex conjugates. Let's take those to be i and -i: another factor is (x²+ 1). The polynomial must be of the form f(x)= (x+F)³(x-1)(x²+ 1). Taking x= 0, f(0)= F³(-1)(1)= -F³, so far. In order that the y intercept be F, we must insure that f(0)= F and we can do that by multiplying the entire polynomial by -F²:
f(x)= -F²(x+F)³(x-1)(x²+1).

4.Create a rational function that has the following characteristics: crosses the x-axis at F; touches the x-axis at -L ; has one vertical asymptote at x=-L-1 and another at x=F+1 ; and has one horizontal asymptote, y=F . Give the formula and a complete graph
Where F = 5, L =4
SOL: Crosses X-Axis @ g
touches X-Axis @ -4
Vertical asymtote X= -4-1 = -5
Veetical asymtote x= 5+1 = 6
Horizontal asymtote y=5
5X 1
--- X ----
(X+5) (X-6)

"crosses the x-axis at F" (does the problem specify that F= 5 or are you assuming that you can choose whatever F you want?) means that there is a factor of (x-F) in the numerator. "touches the x-axis at -L" (I assume "touches" implies that it does not cross the axis) means that there must be a factor of (x-(-L))²= (x+ L)². To have vertical asymptotes at -L-1 and F+1, there must be factors of (x+L+1) and (x-F-1) in the denominator. So far the numerator has degree one more than the denominator. To make sure the horizontal asymptote is F, we must put another x in the denominator- we can do that without introducing another vertical asymptote putting another factor of either (x+L+1) or (x-F-1) in the denominator and then by multiplying the entire fraction by F:
f(x)= F\frac{(x-F)(x+L)^2}{(x-F-1)^2(x+L+1)} is one possible answer.

 

[merikukri]

thanks very much

Yes, you done that one correctly.
Were you given other information you didn't tell us? There is nothing in what you gave here that says h must be any particular function of r.
(And why do you keep saying "F= 5, L= 4"?
You understand, do you not that, that there are an infinite number of correct answers for this? The problem says "Design a polynomial function". You are just asked to come up with one like this. Since the polynomial is to have a zero of degree 3 at x= -F, it must have factors (x-(-F))³= (x+F)³. It has just one more real root, let's make that x= 1, for simplicity: it has a factor of (x-1). The other two roots must be complex conjugates. Let's take those to be i and -i: another factor is (x²+ 1). The polynomial must be of the form f(x)= (x+F)³(x-1)(x²+ 1). Taking x= 0, f(0)= F³(-1)(1)= -F³, so far. In order that the y intercept be F, we must insure that f(0)= F and we can do that by multiplying the entire polynomial by -F²:
f(x)= -F²(x+F)³(x-1)(x²+1).
"crosses the x-axis at F" (does the problem specify that F= 5 or are you assuming that you can choose whatever F you want?) means that there is a factor of (x-F) in the numerator. "touches the x-axis at -L" (I assume "touches" implies that it does not cross the axis) means that there must be a factor of (x-(-L))²= (x+ L)². To have vertical asymptotes at -L-1 and F+1, there must be factors of (x+L+1) and (x-F-1) in the denominator. So far the numerator has degree one more than the denominator. To make sure the horizontal asymptote is F, we must put another x in the denominator- we can do that without introducing another vertical asymptote putting another factor of either (x+L+1) or (x-F-1) in the denominator and then by multiplying the entire fraction by F:
f(x)= F\frac{(x-F)(x+L)^2}{(x-F-1)^2(x+L+1)} is one possible answer.

Thanks very much for your help.

No I wasn't given any other information on cylinderical prob..

and how do I draw a graph for prob 3 & 4. The one with polynomials & Rational questions.

thanks for your help

 

[merikukri]

modified question 2

I am sorry I missed some parts...

Original Question : A Drum in the shape of a right circular cylinder is required to have a volume of 5000 cubic centimeters. The top and bottom are made of material that costs 5 cents per square centimeter, the sides are made of material that cost 4 cents per sq centimeters. (Hint: The formula for the volume of a right circular cylinder is V=Pie r^2 h, where r is the radius of the circular base and h is the height of the cylinder. The surface are of the sides can be determined by cutting the cylinderical shell vertically and flattening it out to get a rectangle whose dimensions can be determined.)

a) Express the total cost C of the material as a function of the radius r of the cylinder.
b) What is the cost if the radius is 25 cm?
c) Graph C=C(r). Using the graph, for what value of r, approximately, is the cost C least?

Sol : C = 4 ( 2pie r h ) + 5 (2pie r ^ 2 )
C(r) = 4 ( 2pie r ( 2 pie r/sroot 3 )) + 5 ( 2pie r^ 2)

2pie r / h = sq root 3 / 1


that's all I soved so far...I am lost, please help !

thanks for ur help