1.A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window find the dimensions of a Norman window of maximum area when the total permeter is 16ft.
2. A rectangle is bounded by the x axis and the semicircle
y=square root 25-x^2. what length and width should the rectangle have so that its area is a maxium.
3. A rectangle package to be sent by a postal service can have a maximum combined length and girth(perimeter of a cross section) of 108 inches. Find the dimensions of the package of maximum volume that can be sent. ( Assume the cross section is square)
Hello to whomever is reading this I do not need the entire problem solved. I just struggle with findind the intial equation. I then know what to do after the equation is found. Can you please help me and let me know if there are any tricks of finding these equations. Thank you
3 answers
Nevermind I got the first one just 2 and three please
I just struggle with findind the intial equation. I then know what to do after the equation is found>>
What you are finding difficult is analysis. You need to learn that skill.
a. make a sketch. Area total=area bottom rectangle + area top semicircle. Take that, and then you have your formula for total area. Of course, perimeter= 2h+1*w*1/2 PI *w check that.
b. make a sketch. You have the formula which the two uppermost corners of the rectangle happen, and with that, you know the x coordinates of the two bottom coordinates.
Rectangle dimensions: 2*xinterceptson curve+2*yinterceptsoncurve
areaRectangle=2x*y where y= 25-x^2
area rectangle= 2x*(25-x^2)
find max area.
What you are finding difficult is analysis. You need to learn that skill.
a. make a sketch. Area total=area bottom rectangle + area top semicircle. Take that, and then you have your formula for total area. Of course, perimeter= 2h+1*w*1/2 PI *w check that.
b. make a sketch. You have the formula which the two uppermost corners of the rectangle happen, and with that, you know the x coordinates of the two bottom coordinates.
Rectangle dimensions: 2*xinterceptson curve+2*yinterceptsoncurve
areaRectangle=2x*y where y= 25-x^2
area rectangle= 2x*(25-x^2)
find max area.
#2 Let the rectangle extend from -x to +x, with height y. Then the area
a = 2xy = 2x√(25-x^2)
now find x such that da/dx = 0
#3 If the square cross-section has side x, and the package has length y, then the girth is 4x, and the volume is
v = x^2y = x^2(108-4x)
a = 2xy = 2x√(25-x^2)
now find x such that da/dx = 0
#3 If the square cross-section has side x, and the package has length y, then the girth is 4x, and the volume is
v = x^2y = x^2(108-4x)
0 answers