Question

alberto is designing a container in the shape of a rectangular prism to ship electronic devices. the length of the container is 10 inches longer than the height. the sum of the length, width, and height is 25 inches. write a function for the volume of the prism. what do the x - intercepts of the graph mean in this context? what dimensions of the container will maximize the volume?
f(x)=-2x^{3}-5x^{2}+150x (simplify your answer)
what are the x - intercepts?
0.75, - 10 (simplify your answer. use a comma to separate answers as needed)
what do the x - intercepts of the graph mean in this context?
the intercepts represent the heights that will result in a container with volume. the intercept is not meaningful because it is not possible to have a height of

Explanation:

Step1: Recall volume formula for rectangular prism

The volume $V$ of a rectangular prism is $V = l\times w\times h$. Given height $h = x$, length $l=x + 10$ and width $w=25$. So $V(x)=25\times x\times(x + 10)$.

Step2: Expand the function

$V(x)=25x(x + 10)=25x^{2}+250x$. But there seems to be an incorrect function $V(x)=- 2x^{3}-5x^{2}+150x$ given in the problem which we will work with for the $x -$ intercept part. To find $x$-intercepts, set $V(x)=0$. So $-2x^{3}-5x^{2}+150x = 0$. Factor out $-x$: $-x(2x^{2}+5x - 150)=0$. Using the quadratic formula for $2x^{2}+5x - 150=0$ where $a = 2$, $b = 5$, $c=-150$, $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-5\pm\sqrt{5^{2}-4\times2\times(- 150)}}{2\times2}=\frac{-5\pm\sqrt{25 + 1200}}{4}=\frac{-5\pm\sqrt{1225}}{4}=\frac{-5\pm35}{4}$. We get $x=\frac{-5 + 35}{4}=\frac{30}{4}=7.5$ and $x=\frac{-5-35}{4}=-10$ and $x = 0$ from $-x=0$.

Step3: Interpret $x$-intercepts

The $x$-intercepts represent the heights ($x$) that result in a container with zero volume. A negative $x$-intercept ($x=-10$) is not meaningful in the context of height (since height cannot be negative). $x = 0$ means the container has no height, so zero volume.

Step4: Maximize the volume

For the correct volume function $V(x)=25x^{2}+250x$, since it is a quadratic function with $a = 25>0$, the vertex of the parabola gives the minimum or maximum. The $x$-coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Here $a = 25$, $b = 250$, so $x=-\frac{250}{2\times25}=- 5$ which is not valid as height cannot be negative. For the given incorrect function $V(x)=-2x^{3}-5x^{2}+150x$, we take the derivative $V^\prime(x)=-6x^{2}-10x + 150$. Set $V^\prime(x)=0$, then $3x^{2}+5x - 75=0$. Using the quadratic formula $x=\frac{-5\pm\sqrt{5^{2}-4\times3\times(-75)}}{2\times3}=\frac{-5\pm\sqrt{25+900}}{6}=\frac{-5\pm\sqrt{925}}{6}=\frac{-5\pm5\sqrt{37}}{6}$. We take the positive value of $x$ (since height cannot be negative) to get the dimension for maximum volume.

Answer:

The volume function should be $V(x)=25x(x + 10)=25x^{2}+250x$. The $x$-intercepts of the incorrect function $-2x^{3}-5x^{2}+150x$ are $x = 0,7.5,-10$. In context, the non - negative $x$-intercepts represent heights that result in a container with zero volume. To find the dimensions for maximum volume, for the correct function, the problem has an issue as the vertex gives a non - valid negative height. For the incorrect function, use the derivative to find the positive $x$ value for maximum volume.