Calc Problem: Find Length & Width of Cheerios Box

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The problem involves finding the length and width of a Cheerios box with a known volume of 182.743 cubic inches, a depth of 2.31245 inches, and a width that is 3.1875 inches less than the length. To solve it, one must set up the volume equation for a rectangular solid, which is length times width times depth. By expressing the width in terms of the length, a quadratic equation can be derived. The discussion highlights confusion over the classification of the problem as a calculus issue, emphasizing that it primarily involves basic geometry and algebra. Ultimately, the focus is on correctly applying the volume formula to find the dimensions of the box.
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The volume of a 10-ounce box of Cheerios is 182.743 cubic inches. The width of the box is 3.1875 inches less then the length, and its depth is 2.31245 inches. Find the length and width of teh box to the thouandanths.

A: I've looked at this problem for a couple of hours, and hunted through my textbook for an equal amount of time, searching for the right formula and how to do it. However, I have been unable to solve this problem.
 
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Can you set up an equation for the volume of the box?
 
If you set the equation up correctly using all the known info (what's the formula for finding the volume of a box), you should wind up with a quadratic equation.
 
You should know that the Volume of a rectangular solid is "length times width times depth". Call the length "l" and then use a "The width of the box is 3.1875 inches less then the length" to write the width as expression in l. You are told that the volume is 182.743 and the depth is 2.31245 so you have one equation in the one unknown l.

This is basic geometry and algebra. Why in the world is it called a "Calculus" problem?
 
It's probably preparation for extrema problems.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

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