Question

a rectangular piece of metal is 25 in longer than it is wide. squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open - box. if the volume of the box is 1800 in³, what were the original dimensions of the piece of metal? what is the original width? in

Explanation:

Step1: Let the original width be $x$ inches.

Then the original length is $x + 25$ inches. After cutting 5 - inch squares from each corner and folding up, the width of the box is $x-10$ inches, the length of the box is $(x + 25)-10=x + 15$ inches, and the height of the box is 5 inches.

Step2: Use the volume formula for a rectangular - box.

The volume $V$ of a rectangular box is $V=l\times w\times h$. We know that $V = 1800$ cubic inches, $l=x + 15$, $w=x - 10$, and $h = 5$. So, $1800=5(x + 15)(x - 10)$.

Step3: Simplify the equation.

First, divide both sides of the equation by 5: $\frac{1800}{5}=(x + 15)(x - 10)$. So, $360=x^{2}-10x+15x-150$.

Step4: Rearrange the equation to a quadratic - form.

Combine like terms: $x^{2}+5x-150 - 360=0$, which gives $x^{2}+5x - 510=0$.

Step5: Solve the quadratic equation.

For a quadratic equation $ax^{2}+bx + c = 0$ (here $a = 1$, $b = 5$, $c=-510$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. First, calculate the discriminant $\Delta=b^{2}-4ac=(5)^{2}-4\times1\times(-510)=25 + 2040=2065$. Then $x=\frac{-5\pm\sqrt{2065}}{2}$. We can also factor the quadratic equation: $x^{2}+5x - 510=(x + 25.5)(x - 20)=0$.
The solutions of the equation $x^{2}+5x - 510=0$ are $x=\frac{-5\pm\sqrt{25+2040}}{2}=\frac{-5\pm\sqrt{2065}}{2}$. Since the width cannot be negative, we consider the positive root. Factoring gives $(x + 25.5)(x - 20)=0$, so $x = 20$ (we ignore $x=-25.5$ as width cannot be negative).

Answer:

20