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? (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 0 and 7.5 represent the heights that will result in a container with zero volume. the intercept - 10 is not meaningful because it is not possible to have a height that is negative. what dimensions of the container will maximize the volume? the function has a maximum when x is approximately . so the container has a maximum volume when height is about 4 in., length is about 14 in., and width is about 7 in. (type whole numbers.)

Explanation:

Step1: Define variables

Let the height of the rectangular - prism be $x$ inches. Then the length is $x + 10$ inches. Let the width be $y$ inches. Given that $x+(x + 10)+y=25$, so $y=25-(2x + 10)=15 - 2x$.

Step2: Write the volume function

The volume $V$ of a rectangular - prism is $V=\text{length}\times\text{width}\times\text{height}$. Substituting the values of length, width, and height, we get $V(x)=(x + 10)(15 - 2x)x=x(15x-2x^{2}+150 - 20x)=-2x^{3}-5x^{2}+150x$.

Step3: Find the 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 $2x^{2}+5x - 150 = 0$ ($x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ with $a = 2$, $b = 5$, $c=-150$) and considering $-x = 0$, the solutions are $x = 0$, $x=\frac{-5\pm\sqrt{25+1200}}{4}=\frac{-5\pm\sqrt{1225}}{4}=\frac{-5\pm35}{4}$. The solutions are $x = 0$, $x=\frac{-5 + 35}{4}=7.5$, $x=\frac{-5 - 35}{4}=-10$. In the context of the problem, $x = 0$ and $x = 7.5$ represent the heights that result in a container with zero volume, and $x=-10$ is not meaningful since height cannot be negative.

Step4: Find the maximum of the volume function

Take the derivative of $V(x)$: $V^\prime(x)=-6x^{2}-10x + 150$. Set $V^\prime(x)=0$, then $3x^{2}+5x - 75 = 0$. Using the quadratic formula $x=\frac{-5\pm\sqrt{25+900}}{6}=\frac{-5\pm\sqrt{925}}{6}=\frac{-5\pm5\sqrt{37}}{6}$. We get $x\approx4$. When $x\approx4$, the length $x + 10\approx14$ and the width $15-2x\approx15 - 8 = 7$.

Answer:

The function for the volume is $V(x)=-2x^{3}-5x^{2}+150x$. The $x$-intercepts $x = 0$ and $x = 7.5$ represent the heights that result in a container with zero volume. The dimensions that maximize the volume are height $\approx4$ inches, length $\approx14$ inches, and width $\approx7$ inches.