What are the dimensions of such a rectangle with the greatest possible area?

Question

A rectangle has its two lower corners on the x-axis and its two upper corners on the parabola y=9-x².

What are the dimensions of such a rectangle with the greatest possible area?

 

1.Width=

2.Height=

Answer 1

This is a parabola that opens vertically, points upward or opens downward and the vertex is located on the Y axis coordinates are:

k=y=9, the vertex is on (0,9), with this data we can work.

The base distance will be equal to 2x at any y less than 9, and the area of the rectangle will be

2x*y = (2*(9-Y)^0.5 )*Y

Now we need to maximize this area using solver to make it faster.

Solver Options
Max Time Unlimited, Iterations Unlimited, Precision 0.000001, Use Automatic Scaling
Convergence 0.0001, Population Size 100, Random Seed 0, Derivatives Forward, Require Bounds
Max Subproblems Unlimited, Max Integer Sols Unlimited, Integer Tolerance 1%, Assume NonNegative
Objective Cell (Max)
Cell Name Original Value Final Value
$E$3 5.656854249 20.78
The answer is: height (Y) = 6 units and then x will be known with the parabola equation, will be X= 3.47

and the area (max) will be 20.78

Variable Cells
Cell Name Original Value Final Value Integer
$E$4 Y 1 5.99999989 Contin


Constraints
Cell Name Cell Value Formula Status Slack
$E$4 Y 5.99999989 $E$4<=9 Not Binding 3.00000011

 

 

Answer 2

Hi,

 

When I was in school, Bill Gates was just born, so we did not have Excel yet. In other words, if you wanted a wrong answer, it took a lot longer to get it; today, just open Excel, and you can be wrong really fast.

 

So, I will solve this the "old-fashioned way". We let "x" be half the base of the rectangle -- the base goes from -x to +x, totaling 2x in length. The height is "y", and y is on the parabola, y = 9 - x². Therefore the area of the rectangle is

 

   A = 2x * y

      = 2x * (9 - x²)

      = 18x - 2x³

 

And now the old-fashioned math: to find the maximum, take the derivative, set it =0, solve, and make sure the solution is a maximum not a minimum.

 

Thus, the derivative dA/dx = 18 - (2 * 3 * x²) = 18 - 6x²

 

Setting dA/dx=0 gives us

 

     18 - 6x² = 0

so, 18 = 6x²

so, 3 = x²

so, x = ±√3

 

We have defined "x" as the length of half the base, so our solution is the positive root, x=√3. To check if this solution is a max/min, we take the second derivative -- d²A/dx² = -2*6*x¹ = -12x. Therefore, at x=√3, the second derivative is d²A/dx² = -12√3, which is negative. Therefore our solution is a maximum, and we have solved the problem. Note that for x=√3, y = 9-x² = 6

 

To summarize, the maximum area is at (x,y) = (√3, 6), corresponding to a rectangle with base of 2x = 2√3, and height of 6. The area is (2√3 * 6) = 12√3 ≈ 20.78

 

Wow - we solved it without Excel, and got the correct answer.

 

I hope this has helped you out.

Answer 3

Width = 6

 

Height = 9

 

Area = 9*6= 54