Mathematics, 13.08.2021 20:50 mackenziewright2002

1 A metallic bar of length 96 inch was used to make a rectangular frame. Find the dimension of the rectangular metallic frame with maximum area.

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Answer from: tedrayoung1

The dimensions of the rectangle that maximizes the area is 24 inches by 24 inches.

Step-by-step explanation:

We are given that a metallic bar with a length of 96 inches was used to make a rectangular frame.

And we want to find the dimensions of the frame that yields the maximum area.

Since the bar has a length of 96 inches, the perimeter of the rectangle must total 96. Let l represent the length of the rectangle and w its width. Hence:

$96 = 2\left( \ell + w \right)$

Divide:

$\displaystyle \ell + w = 48$

The area of the rectangle will be:

$\displaystyle A = w\ell$

From the first equation, solve for either variable:

$\displaystyle w = 48 -\ell$

Substitute:

$\displaystyle A = \ell (48 -\ell)$

Distribute:

$\displaystyle A = 48\ell - \ell^2$

Since the area of the rectangle is represented with a quadratic equation with a negative leading coefficient, the maximum area will occur at its vertex point.

The vertex of a quadratic is given by:

$\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$

In this case, a = -1, b = 48, and c = 0.

Then the length for which the maximum area occurs is:

$\displaystyle \ell = -\frac{(48)}{2(-1)} = 24$

And the width will be:

$\displaystyle w = 48 - \left(24\right) = 24$

In conclusion, the dimensions of the rectangle that maximizes the area is 24 inches by 24 inches.

Answer from: ayoismeisjjjjuan

$\displaystyle 24\text{ in x }24\text{ in, achieving a maximum area of } 576\:\mathrm{in^2}$

Step-by-step explanation:

The general concept to solve this problem is to write the area of the rectangle as a quadratic, forming a parabola with some maximum value at the vertex. That maximum value will represent the maximum possible area of the rectangle.

Let represent the width of the rectangle and represent the length of the triangle.

Since the perimeter of the rectangle is 96 inches, we have:

2x+2y=96

The area of a rectangle with width and length $\ell$ is given by $A=w\ell$ . Therefore, the area of the rectangle with width and length is equal to A=xy .

Isolate in the first equation 2x+2y=96 so that we can substitute it into A=xy :

$\\2x+2y=96,\\2y=96-2x,\\y=\frac{96-2x}{2}=48-x$

Now substitute y=48-x into A=xy :

$A=x(48-x),\\A=48x-x^2,\\A=-x^2+48x$

This is a parabola concave down. The y-coordinate of the vertex represents the maximum area of the rectangle and the x-coordinate of the vertex represents the width to achieve that area.

The x-coordinate of the vertex of a parabola in standard form ax^2+bx+c is given by $\frac{-b}{2a}$ .

In -x^2+48x , assign values:

$a\implies -1$ $b\implies 48$ $c\implies 0$

Therefore, the x-coordinate of the vertex of -x^2+48x is equal to:

$\frac{-48}{2(-1)}=\frac{-48}{-2}=24$

Now substitute x=24 into -x^2+48 to find the y-coordinate of the vertex:

$A=-(24)^2+48(24),\\A=-576+1152,\\A=\boxed{576\:\mathrm{in^2}}$

Furthermore, we find that the dimensions of the rectangle to achieve the maximum area possible must be 24 in x 24 in.

Answer from: Quest

i pretty sure it is $\frac{9}{500}$

Answer from: Quest

you could deduce the conclusion directly using the facts pertaining to the specific situation.

hope this .

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1 A metallic bar of length 96 inch was used to make a rectangular frame. Find the dimension of the r...