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The height is just the size of the corner cut out ( x in this problem). The length and width of the bottom of the box are both smaller than the cardboard because of the cut out corners. So the volume, as a function of x, is given by V ( x) = x (25 - 2 x ) (20 - 2 x ).
- Extreme Value Theorem
EVT doesn't guarantee uniqueness of global extrema, just...
- L'hopital's Rule
when f (a) = g (a) = 0.One method is to use L'Hopital's...
- Polar Derivatives
Polar Derivatives. Curves can also be defined using polar...
- Related Rates
1. Initial approach. The applet shows an image of the...
- Basic Curve Analysis
Exrema, maxima and minima. Critical points are candidates...
- Global Extrema
Now there are three local extrema, two minima and one...
- Motion on a Plane
Motion on a Plane. We can use calculus to understand the...
- Motion Along a Line
Motion along a line. We can use calculus to understand the...
- Extreme Value Theorem
- Step 2: Create Your Objective Function and Constraint Equation
- Step 3: Rearrange Constraint Equation and Substitute Into Objective Function
- Step 4: Take The Derivative, Set It Equal to Zero, and Solve
- Step 5: Check If The Point X = 8 Is A Max, Min, Or Saddle Point
- Step 6: Solve For Y
The objectivefunction is the formula for the volume of a rectangular box: \[ V = \text{length} \times \text{width} \times \text{height} = X \times X \times Y \\[2ex] V = X^2Y\] The constraintequation is the total surface area of the tank (since the surface area determines the amount of glass we'll use). The surface area is simply the sum of the are...
\[ V = X^2 \times Y \\[2ex] SA = X^2 + 4XY = 192 \\ \quad \quad \quad - X^2 \quad \quad - X^2 \\[2ex] 4XY = 192 - X^2 \\[2ex] \frac{4XY}{4X} = \frac{192 - X^2}{4X} \\[2ex] Y = \frac {1}{4X} \times (192-X^2) \\[2ex] V = X^2 \times \frac{1}{4X} \times (192-X^2) \\[2ex] V = \frac{X^2}{4X} \times (192-X^2) \\[2ex] V = \frac{X}{4} \times (192-X^2) \\[2e...
\[ V' = \frac{1}{4} \times (192 - 3X^2) = 0 \\[2ex] V' = [\frac{1}{4} \times (192 - 3X^2)] \times 4 = 0 \times 4 \\[2ex] V' = 192 - 3X^2 = 0 \\ \phantom{V' = } - 192 \quad \quad - 192 \\[2ex] - 3X^2 = -192 \\[2ex] \frac{-3X^2}{-3} = \frac{-192}{-3} \\[2ex] X^2 = 64 \\[2ex] \sqrt{X^2} = \sqrt{64} \\[2ex] X = 8\]
Test slope at \(X=7\) \[ V'(7) = \frac{1}{4} \times (192 - 3(7)^2) \\[2ex] V'(7) = \frac{1}{4} \times (192 - 147) \\[2ex] V'(7) = \frac{1}{4} \times 45 \\[2ex] V'(7) = \frac{45}{4} = 11.25 \\[2ex]\] Test slope at \(X=9\) \[ V'(7) = \frac{1}{4} \times (192 - 3(9)^2) \\[2ex] V'(7) = \frac{1}{4} \times (192 - 243) \\[2ex] V'(7) = \frac{1}{4} \times -5...
\[ Y = \frac {1} {4X} \times (192-X^2) \\[2ex] Y = \frac {1} {4 \times 8} \times (192-8^2) \\[2ex] Y = \frac {1} {32} \times (192-64) \\[2ex] Y = \frac {1} {32} \times (128) \\[2ex] Y = \frac {128} {32} \\[2ex] Y = 4\] An \(8 \times 8 \times 4\)inch tank gives us the maximum volume.
Jul 21, 2018 · In order to maximise the volume, the area enclosed by the rectangle, $l\times w$ should be maximum and thus, $l=w$. Therefore we may write: $2w=62-h$, or, $w=31-\frac1 2 h$. Now substitute this in the equation for total volume, $V=lwh=(31-\frac 1 2 h)^2h$.
Sep 24, 2020 · What is the maximum volume (or capacity) of the box, and what are the dimensions of this maximum-volume box? Here is the reinterpreted problem; observe that the meaning of the variables has changed: Now we’ll have a considerably more complicated formula for the area that is 12 m 2 , but the volume formula is far simpler.
Nov 10, 2020 · In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain volume. In this section, we show how to set up these types of minimization and maximization problems and solve them by using the tools developed in this chapter.
Continuum mechanics. In nonideal fluid dynamics, the Hagen–Poiseuille equation, also known as the Hagen–Poiseuille law, Poiseuille law or Poiseuille equation, is a physical law that gives the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of constant cross section.
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Mar 26, 2016 · You’ve got your answer: a height of 5 inches produces the box with maximum volume (2000 cubic inches). Because the length and width equal 30 – 2 h, a height of 5 inches gives a length and width of 30 – 2 · 5, or 20 inches.
