Consider a window the shape of which is a rectangle of height h surmounted a triangle having a height T that is 1.3 times the width w of the rectangle.
If the cross-sectional area is A, determine the dimensions of the window which minimize the perimeter.
h=______
w=______
What I did was:
Area:
Rectangle: h*w
Triangle: w * 1.3w / 2 = 0.65w^2
The complete area: hw + 0.3w^2.
Perimeter:
3 sides of the rectangle: 2h+w
Twice the sloped side of the triangle: s^2 = (w/2)^2 + (1.3w)^2 = w^2(0.25+1.69)=1.94w^2, so s = 1.39w.
The complete perimeter p = 2h+w+2.78w = 2h+3.78w
A = hw + 0.65w^2
A - 0.65w^2 = hw
A/w - 0.65w = h
p = p(w) = 2h + 3.78w
= 2(A/w-0.65w) + 3.78w
= 2A/w - 1.3w + 3.78w
= 2A/w + 2.48w
p'(w) = (-1)2A/w^2 + 2.48
p'(w_min) = (-1)2A/w_min^2 + 2.48 = 0
-2A + 2.48w_min^2 = 0
w_min^2 = A/1.24
w_min = sqrt(A/1.24)
So the dimensions I got are:
w_min = sqrt(A/1.24)
h_min = A/w_min - 0.65w_min = A/sqrt(A/1.24) - 0.6sqrt(A/1.24) =
= sqrt(1.24A)-0.65sqrt(A) = sqrt(A) [sqrt(1.24)-0.65].
Calculus (pleas help I really need help with this) - MathMate, Monday, April 11, 2011 at 12:01am
There's this line:
The complete area: hw + 0.3w^2.
which should read hw+0.65w^2.
But I think your subsequent calculations use the correct expression.
Your approach appears correct, although I did not check the arithmetic.
That is one problem with computerized exercises.
Were there instructions as to how you present the results, such as:
- in decimals to two decimal places, or
- in fractions
- exact expression in fractions, or
any other instructions.
If the question requires an accuracy to 2 decimal places, I would carry all calculations to 4 places until the last, when I enter only two places, probably rounded.
Another possible source of problem is the interpretation of the "cross section" area. I have not heard of a cross section area of a window. Does it refer to the whole window, as you did, or just the rectangular part?
Finally, the last expression should read:
sqrt(A) [sqrt(1.24)-0.65/�ã(1.24)].
Calculus (pleas help I really need help with this) - Katrina, Monday, April 11, 2011 at 12:20am
I tried that... it's not it... apparently there must be something wrong with the arythmetic but I can't find what it is...
Calculus (pleas help I really need help with this) - Katrina, Monday, April 11, 2011 at 12:55am
The cross-sectional area is jus the rectangle...
Calculus (pleas help I really need help with this) - MathMate, Monday, April 11, 2011 at 7:33am
Does that mean that you've got the right answer?
I reposted this because it's so far away now... I didn't get the right answer, can someone please check what I have wrong?
Did you redo your calculations based on A=w*h?
If you did, you can post what you've got.
Also, how many "lives" do you have left?
A=wh
h=A/w
Perimeter,
P(w)=2h+w+2√((w/2)^2+(1.3w)^2)
=2A/w + (1+(√776)/10)w
P'(w)=2A/w^2 + 1+ sqrt(194)/5
Set P'(w)=0 and solve for w to get:
w=±(sqrt(10A))/sqrt(sqrt(194)+5)
=0.727√A (reject negative root)
To determine the dimensions of the window that minimize the perimeter, we can use calculus. Let's go through the steps again:
1. Define the variables:
- h: height of the rectangle
- w: width of the rectangle
- T: height of the triangle
- A: cross-sectional area of the window
- p: perimeter of the window
2. Express the area and perimeter in terms of h and w:
Area:
Rectangle: A_rect = h * w
Triangle: A_tri = (1/2) * w * T = (1/2) * w * 1.3w = 0.65w^2
Complete area: A = A_rect + A_tri = h * w + 0.65w^2
Perimeter:
3 sides of the rectangle: p_rect = 2h + w
Twice the sloped side of the triangle: p_tri = 2 * sqrt((w/2)^2 + (1.3w)^2) = 2 * sqrt(0.25w^2 + 1.69w^2) = 2 * sqrt(1.94w^2) = 2 * 1.39w = 2.78w
Complete perimeter: p = p_rect + p_tri = 2h + w + 2.78w = 2h + 3.78w
3. Write the area equation in terms of h only:
A = hw + 0.65w^2
h = A/w - 0.65w
4. Write the perimeter equation in terms of h only:
p = 2h + 3.78w
5. Find the derivative of the perimeter with respect to w:
dp/dw = 0 (since the perimeter is minimized when its derivative is equal to zero)
6. Solve for the value of w that minimizes the perimeter:
dp/dw = 0
0 = 2h' + 3.78
3.78 = -2h'
h' = -3.78/2
7. Substitute h' into the expression for h in terms of w:
h = A/w - 0.65w
h = A/w - 0.65(-3.78/2)
h = A/w + 1.23w
8. Now we can solve for the dimensions of the window:
From the area equation: h = A/w + 1.23w
Plug in the expression for h from the perimeter equation: h = -3.78/2 + 1.23w
Solve for w:
-3.78/2 + 1.23w = A/w + 1.23w
-3.78/2 = A/w
Multiply both sides by 2w:
-3.78w = 2A
Solve for w:
w = -2A/3.78
Now substitute w back into the equation for h:
h = A/(-2A/3.78) + 1.23(-2A/3.78)
h = -3.78/2A + (-2A/3.78)(1.23)
h = -3.78/2A - 2A(1.23)/3.78
h = -3.78/2A - 2.46A/3.78
The dimensions of the window that minimize the perimeter are:
h = -3.78/2A - 2.46A/3.78
w = -2A/3.78
Note that the negative sign in the solution for h and w indicates that the dimensions are imaginary. There may have been an error in the previous calculations or in the problem statement. Please double-check your calculations and the given information to ensure an accurate solution.
To determine the dimensions of the window that minimize the perimeter, we need to find the critical points of the function representing the perimeter and then determine which critical point corresponds to the minimum perimeter.
Let's start by finding the derivative of the perimeter function with respect to the width, w:
p'(w) = -2A/w^2 + 2.48
Now, set p'(w) equal to 0 and solve for the width, w:
-2A/w^2 + 2.48 = 0
2A = 2.48w^2
w^2 = A/1.24
w = sqrt(A/1.24)
So, the width of the window that minimizes the perimeter is w = sqrt(A/1.24).
To find the height, h, substitute this value of w into the expression for h:
h = A/w - 0.65w
h = A/sqrt(A/1.24) - 0.65sqrt(A/1.24)
Simplifying:
h = sqrt(1.24A) - 0.65sqrt(A)
Therefore, the dimensions of the window that minimize the perimeter are:
w = sqrt(A/1.24)
h = sqrt(1.24A) - 0.65sqrt(A)
Please note that there might be arithmetic errors in your calculations that have led to a different result. Double-check your calculations to ensure accuracy.