Chapter 1.5, Problem 88E

i.

To determine

To calculate: Corners of equal size are cut from a square with sides of length 8 meters. Write the area A of the function of x. Determine the domain of the function.

  

Answer to Problem 88E

Area of the resulting figure is $A\left(x\right)=64-2x^2,x\in \left[0,4\right]$

Explanation of Solution

Given information: size of square is 8 meters.

Area of the resulting figure is the difference between the area of square of side 8 meters and the 4 right angled isosceles triangle of side x.

Calculation:

Area of square is $8^2=64$ and area of 4 right angled isosceles triangles is $4\times \frac{1}{2}\times x\times x=2x^2$

So resulting area is $A\left(x\right)=64-2x^2$

For the given data to make sense $2x\le 8\Rightarrow x\le 4$ and x is length so $x\ge 0$

  $\Rightarrow x\in \left[0,4\right]$

Conclusion: Area of the resulting figure is $A\left(x\right)=64-2x^2,x\in \left[0,4\right]$

ii.

To determine

To graph: Corners of equal size are cut from a square with sides of length 8 meters. Use a graphing utility to graph the area function over its domain. Use the graph to find the range of the function.

Answer to Problem 88E

  

Explanation of Solution

Between $\left[0,4\right]$ the function decrease

The given function is $A\left(x\right)=64-2x^2$ keeping values of x in this equation we will get the graph.

ii.

To determine

To calculate: Corners of equal size are cut from a square with sides of length 8 meters. Identify the figure that results when x is the maximum value in the domain of the function. What would be the length of each side of the figure?

Answer to Problem 88E

The remaining figure is Square with length of each 5.65 meters.

Explanation of Solution

Given information: Side of given square is 8 meter and $2x\le 8$ so let maximum value of $2x=8\Rightarrow x=4$

Now two sides of right angled isosceles triangle is 4 meters.

Calculate the hypotenuse of the right angled triangle with two side x meter each.

Length of hypotenuse $=\sqrt{4^2+4^2}=4\sqrt{2}$

Conclusion: each side of resulting figure is $4\sqrt{2}$ and both diagonals are of same length so it is square.