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
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.
#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)
Certainly! I can help you with finding the initial equations for these optimization problems.
1. Norman Window:
Let's define the dimensions of the rectangular window as length (L) and width (W), and the radius of the semicircle as r. The total perimeter is given as 16ft, so we have the equation:
2L + W + πr + 2r = 16.
To maximize the area, we need to express the area A in terms of a single variable. The area of the rectangular part is A_rect = L * W, and the area of the semicircle is A_semi = 1/2 * πr^2. So, the total area is A = A_rect + A_semi = L * W + 1/2 * πr^2.
Now, we can express one variable in terms of the other using the given perimeter equation. For example, we can solve for L in terms of W and r: L = (16 - W - πr - 2r) / 2.
Substituting this back into the area equation, we get A = [(16 - W - πr - 2r) / 2] * W + 1/2 * πr^2.
To find the maximum area, you can take the derivative of A with respect to either W or r, set it equal to zero, and solve for the other variable. Then substitute the values back into the area equation to find the maximum area.
2. Rectangle and Semicircle:
Let's assume the width of the rectangle is 2x, and the length of the rectangle is y. The equation of the semicircle is y = √(25 - x^2).
The area of the rectangle is A_rect = 2x * y = 2xy.
To maximize the area, we need to express the area A in terms of a single variable. So, A = 2xy.
Now, we need to eliminate either x or y from the equation by substituting using the equation of the semicircle. To do that, we can square both sides of the equation to get rid of the square root: y^2 = 25 - x^2.
Now, we can substitute y^2 in the area equation: A = 2x * (25 - x^2).
To find the maximum area, you can take the derivative of A with respect to x, set it equal to zero, and solve for x. Then substitute the value back into the area equation to find the maximum area.
3. Rectangle Package:
Let's assume the dimensions of the square cross-section are both x inches. The length of the package would be the sum of the square cross-section perimeter and twice the width x: 2x + 2(x) = 2x + 4x = 6x.
The combined length and girth should not exceed 108 inches, so we have the equation:
6x ≤ 108.
To find the maximum volume, we need to express the volume V in terms of a single variable. The volume of the rectangular box is V_box = x * x * 6x = 6x^3.
Now, we can substitute 6x from the perimeter equation into the volume equation: V = (108/6)^3 = 18^3 = 5832 cubic inches.
Please note that for these optimization problems, you may want to verify that the critical points you find correspond to maximum values by checking the second derivative or simply using common sense.
I hope this helps you to find the initial equations for each problem. If you have any further questions, feel free to ask!