pogonofor9z

2021-12-17

A certain type of candle is packaged in boxes that measure 42 cm by 17 cm by 10 cm. The candle company that produced the above packaging has now designed shorter candles. A smaller box will be created by decreasing each dimension of the larger box by the same length. The volume of the smaller box will be at the most 2220 cm3. What are the maximum dimensions of the smaller box? Solve this problem algebraically. Please use The Factor Theorem

Cleveland Walters

Expert

Step 1
Given : A candle is packaged in boxes that measure 42cm by 17cm by 10cm .
The volume of the smaller box $=2220c{m}^{3}$
Step 2
Proof: Let the decreasing dimension of the larger box for the smaller box be x
So, the dimension of smaller box will become .
Volume of the box $=l×b×h$
$\left(42-x\right)\left(17-x\right)\left(10-x\right)=2220$
$-{x}^{3}+69{x}^{2}-1304x+7140=2220$
$-{x}^{3}+69{x}^{2}-1304x-2220=2220-2220$
$-x3+69{x}^{2}-1304x+4920=0$
$-\left(x-5\right)\left(x2-64x+984\right)=0$

$x=5,x=2\left(16+\frac{101}{2}\right),x=2\left(16-\frac{101}{2}\right)$
The value of $x=2\left(16+\frac{101}{2}\right),x=2\left(16-\frac{101}{2}\right)$ cannot be considered because will be negative .
Therefore, $x=5$ will be the maximum value so the dimension of smaller box will be 37cm by 12cm by 5cm .

einfachmoipf

Expert

Explanation:
As given the candle is packed in the boxes that measure $42cm×17cm×10cm$ and smaller box is created by decreasing each dimension of the larger box by the same length.
So, consider the decreased length is "x" cm.
Hence, the dimension of the smaller box will be:

we also have the volume of the smaller box which is $2220c{m}^{3}$
and since, the volume of the rectangular box $=l×b×h$
$⇒\left(42-x\right)\left(17-x\right)\left(10-x\right)=2220$
So, let $f\left(x\right)=\left(42-x\right)\left(17-x\right)\left(10-x\right)-2220=0$ (i)
Now, we will check the different values of x for which $f\left(x\right)=0$:
$f\left(x\right)=\left(42-x\right)\left(17-x\right)\left(10-x\right)-2220=0$
$f\left(1\right)=\left(42-1\right)\left(17-1\right)\left(10-1\right)-2220=\left(41\right)\left(16\right)\left(9\right)-2220=3684\ne 0$
$f\left(2\right)=\left(42-2\right)\left(17-2\right)\left(10-2\right)-2220=\left(40\right)\left(15\right)\left(8\right)-2220=2580\ne 0$
$f\left(3\right)=\left(42-3\right)\left(17-3\right)\left(10-3\right)-2220=\left(39\right)\left(14\right)\left(7\right)-2220=1602\ne 0$
$f\left(4\right)=\left(42-4\right)\left(17-4\right)\left(10-4\right)-2220=\left(38\right)\left(13\right)\left(6\right)-2220=744\ne 0$
$f\left(5\right)=\left(42-5\right)\left(17-5\right)\left(10-5\right)-2220=\left(37\right)\left(12\right)\left(5\right)-2220=0$
$\therefore$ the decreased length is $x=5cm$
Hence, the dimensions of smaller box will be:
$42-5=37cm$
$17-5=12cm$
$10-5=5cm$
Answer: The dimensions of smaller box is:
37cm, 12cm, and 5cm.

Don Sumner

Expert

l=42cm, b=17cm, h=10cm
The dimensions decreased by x volume of new box $=2220c{m}^{3}$
$\left(l-x\right)\left(b-x\right)\left(4-x\right)=2220$
$\left(42-x\right)\left(17-x\right)\left(10-x\right)=2220$
$\left(x-42\right)\left(x-17\right)\left(x-10\right)+2220=0$
${x}^{3}-69{x}^{2}+1304x-4920=0$
x=5, 25.68, 38.33
Dimension of smaller box will be max at =5
37cm, 12cm, 5cm

Do you have a similar question?