Find the dimensions of the rectangle with the largest area

Question

Find the dimensions of the rectangle with the largest area; it has two vertices on the x-axis and two vertices above the x-axis on the graph of y = 14 − x².

length of the horizontal side

length of the vertical side

I've tried this a few times and my answers are the same, but it is wrong

for the horizontal side, i got 4.32

for vertical, i got 9.33

Both these answers are wrong.

Answer 1

The parabola has symmetry with respect to the y axis.

Let (x, 14-x²) be the upper right corner of the rectangle.

Area of rectangle = A = (2x)(14 - x²) = 28x - 2x³

A' = 28 - 6x²

A' = 0 when x² = 14/3

Horizontal side = 2√(14/3) and Vertical side = 14 - 14/3 = 28/3

Answer 2

First of all if the two vertices fall on the X-axis and it is a rectangle, then the Y should be identical.

Let Y₁ = 14-X₁²

Y₂ = 14-X₂²

^{But Y1 = Y2}

^=> 14-X₁² = 14-X₂²

14-14 -X₁² = -X₂²

X₁² = X₂²

X₁ = ±√(X₂²)

X₁ = ±X₂

Side length = X₂ - X₁

If X₁ = X₂ Side Length would be zero and This implies that the Xs are equal but opposite in sign.

there fore X₁ = -X₂

Side length = X₂ - -X₂ = X₂ + X₂ = 2X₂

_Since X₁ = -X₂

Y₂ = 14-X₂²

Area = Y_{2 *} Side Length

Area = Y_{2 *} 2X₂

Substituting Y₂

Area = (14-X²) _* 2X₂

For simplicity use X instead of X₂ and A instead of Area

Area = (14-X²) _* 2X

Area = 28X-2X³

The derivative of this is:

dA/dx = 28-6X²

To get the maximum set dA/dx =0

dA/dx = 28-6X² = 0

X = ±√(28/6) = ±√(14/3)

take X = √(14/3)

Substituting,

Area max = 28X-2X³ = √(14/3)*[28 - 2* (√(14/3)²]

Area max = √(14/3)*[28 - 2* 14/3]

Area max = √(14/3)*[56/3]

Side Length is 2X₂ = 2*√(14/3)

To find the vertical side:

Area max/Side Length = (√(14/3)*[56/3] )/(2*√(14/3)) = 28/3

OR

Y2 = Y1 = Y = 14-X₂² = 14- (√(14/3))² = 28/3