Q1.A rectangular storage box has a base that is three times as long as
it is wide.  The top is made of a material that costs 35 times the
price per square metre of the material used for the rest of the box. 
Find the dimensions that would be necessary to ensure the minimum cost
if the volume of the box must be 3 m^3.

Q2. A window is the shape of a rectangle with an equilateral triangle
on top.  The total perimeter of the window is 300 cm.  Find the width
that will allow the maximum amount of light to enter.  Note: In an
equilateral triangle, height=Sq.Rt.3/2 x length of each side.

Hi salimjavaid!
Here are the answers to your questions:

Q1. The problem here is to find the sides of the box that minimize its
cost subject to the constraint that the volume of the box should be
equal to 3 m^3.

Let's call 'x' the length of the short side of the rectangular base
(measured in meters). So the length of the other side would be 3x.
Let's call 'y' the height of the box. From here, we can write the
volume constraint as:

3x * x * y = 3
3x^2 * y = 3
x^2 * y =1

Now let's write the cost function. The area of the base (equal to the
area of the top) is 3x*x. The area of two of the sides is x*y. The
area of the other two sides is 3x*y. All the sides of the box, cost 1
per square meter, while the top costs 35 per square meter. Then, the
cost of the box is:

Cost of base = 3x * x * 1 = 3x^2
Cost of 2 of the sides = 2 * x * y * 1 = 2xy
Cost of other 2 sides = 2 * 3x * y * 1 = 6xy
Cost of top = 35 * 3x^2 = 105x^2

Therefore, the total cost of the box is

Cost =  3x^2 + 2xy + 6xy + 105x^2
     = 108x^2 + 8xy

So the problem is:

Min      108x^2 + 8xy
x,y    subject to x^2*y=1

We can replace the constraint into the cost function to obtain:

Cost = 108x^2 + (8x)*(1/x^2)
     = 108x^2 + 8/x

So we have to choose x such that this cost function is minimum. We can
do this simply be taking the derivative of this function with respect
to x and equating it to zero. We get:

Cost' = 216x - 8/(x^2)

So we must solve

216x - 8/(x^2) = 0
216x = 8/x^2
216x^3 = 8
x^3 = 8/216
x = 1/3

So now we have the answer. One of the sides of the base should be 1/3
m long. The other side should be three times that: 1 m. What about the
height? We can solve for it using the constraint equation now that we
know x:

x^2 * y = 1
(1/3)^2 * y = 1
y = 9

Therefore, the height of the box should be 9 m.


Q2. Let's draw the windows in order to better understand this problem:

    /\
  x/  \x
  /    \
 /      \
 |      |
 |      |y
 |      |
 |______|
    x

We have to find here the length 'x' such that the area of this window
is maximum (maximum amount of light possible), subject to the
constraint that the perimeter of the window whould be 300 cm.

Let's write the constraint equation. From the rectangle part, we have
3 sides: 1 of length 'x' and 2 of length 'y'. From the triangle part
we have 2 sides, of length 'x'. Therefore, the perimeter is 3x+2y. So
the constraint would be:

2x + 3y = 300

Now let's write the function we want to minimize. The area of the
rectangle is xy. The area of the triangle is (1/4)*sqrt(3)*x^2.
Therefore, the total area of the window is

Area = xy + (1/4)*sqrt(3)*x^2.

Areas, Volumes and Surfaces
http://www.math2.org/math/geometry/areasvols.htm

So the problem can be written as

Max    xy + (1/4)*sqrt(3)*x^2
x,y   subject to 3x + 2y = 300

Again, we replace the constraint into the area function in order to
leave only 'x' in it. From the constraint:

y = 300/2 - 3x/2

Replacing this into the area function:

Area = x*300/2 - x*3x/2 + (1/4)*sqrt(3)*x^2
     = 150x - 3(x^2)/2 + (1/4)*sqrt(3)*x^2
     = 150x + x^2*( sqrt(3)/4 - 3/2)
     = 150x + x^2*( sqrt(3)/4 - 6/4)
     = 150x + x^2*(sqrt(3)-6)/4

We can find the maximum by taking the derivative of this function and
equating it to 0:

Area' = 150 + x*(sqrt(3)-6)/2

So,

0 = 150 + x*(sqrt(3)-6)/2
-150/(sqrt(3)-6) = x/2
x = 300/(6-sqrt(3))
x = 70.29...

Therefore, the width of the window should be 70.29 cm. Again, you can
calculate 'y' using the value of x and the constraint equation.


Note that in both questions, you actually need to find the 2nd
derivative (the derivative of the derivative) to confirm that you have
actually find either a minimum (in the first question) or a maximum
(in the second question). When the 2nd derivative is positive, the
function is called "convex", and the result from equating the 1st
derivative to 0 (the procedure we used) is the minimum. When the 2nd
derivative is negative, the function is called "concave", and the
result from equating the 1st derivative to 0 is the maximum. You'll
find that in Question 1, the 2nd derivative turns out to be positive,
while in Question 2 it's negative, confirming our results in both
cases.


I hope this helps! If you have any doubts regarding my answer, please
don't hesitate to request clarification before rating it.

Best wishes!
elmarto