thanuellekare
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Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola.
y=5-x^2

Respuesta :

meerkat18
let 2x be the length of rectangw where x is value of x of point on parabola width is represented as y is the length.

  Area = 2x*y = 2x (5-x^2) = 10x -2x^3
 maximize Area by finding x value where derivative is zero
 dA/dx = 10 -6x^2 = 0
 --> x = sqrt(5/3)

optimal dimensions: length = 2sqrt(5/3) width = 10/3
abidemiokin

The dimension of the rectangle is √5/3 by 40/9

From the question shown,

Let 2x be the length of the rectangle where;

x is the value of any point on the parabola

y is the width of the triangle;

Area of the rectangle = length * width

A = 2xy

Given the equation of the parabola expressed as y = 5-x²

A = 2x(5-x²)

A = 10x - 2x³

If the area of the rectangle is maximized, dA/dx = 0

10 - 6x² = 0

6x² = 10

x² = 10/6

x² = 5/3

x = ±√5/3

x =√5/3

Since  

y = 5 - (√5/3)²

y = 5 - 5/9

y = 40/9

Hence the dimension of the rectangle is √5/3 by 40/9

Learn more here: https://brainly.com/question/6568399

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