Advanced Placement (AP), 16.01.2020 14:31 blackbetty79

If 10,800 cm2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

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What are you given variables?
Surface area of box: 10800cm²
Volume of box: s²h
Where:
SA=surface area
s = side of square base
h = height of box

Make the height of the box in terms of s
You can write the formula for the surface area of the box in terms of s and h like so:
S.A. = s² + 4sh
Where: S.A. = surface area or 1200 cm², s² = the square base, and 4sh = the four 'walls' of the box.
10800 = s² + 4sh
10800 - s² = 4sh
(10800 - s²)/(4s) = h

Substitute h (in terms of s) into the formula for volume.
v(s) = s²((10800- s²)/(4s)) // Simplify.
v(s) = s(10800 - s²)/4 // Expand.
v(s) = 2700s - (1/4)s^3 // To find the largest possible volume of the box, you find the maximum value of this function.

Take the derivative of the volume function using the Power Theorem.
v'(s) = 2700 - (3/4)s² // Zeroes of the d/dx v function will give you the x values that correspond to local extrema in the v function.
0 = 2700 - (3/4)s² // Solve for zeroes.
-2700= (-3/4)s²
3600= s²
Your two zero values for s are -60 and 60.

Take the second derivative to see if the s values will give you a local maximum or minimum.
v"(s) = -(3/2)s
v"(-60) = -(3/2)(-60)
v"(-60) = 90 // This indicates a local minimum for v at s, not what we are looking for.
v"(60) = -(3/2)(60)
v"(60) = -90 // This indicates a local maximum for v at s, which is what we are looking for, the maximum volume of the box. This makes sense because if you remember what we assigned the variable s to, a side length, side lengths cannot be negative.

Once we have found the value for s, we can substitute it into the function we created for the volume of the box.
v(s) = 2700s- (1/4)s^3
v(s) = 2700(60) - (1/4)(60)^3
v(s) = 162000- (1/4)(216000)
v = 162000- 54000

The largest possible volume of the box is 108000 cubic centimeters.

Answer from: Quest

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Answer from: Quest

Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

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