Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

The surface area of a cube is increasing at a rate of 15 square meters per hour. At a certain instant, the surface area is 24 square meters. What is the rate of change of the volume of the cube at that instant (in cubic meters per hour)? Choose 1 answer: (A) 8 (B) (15)/(2) (C) (sqrt15)^(3) (D) (5)/(8)

The surface area of a cube is increasing at a rate of 1515 square meters per hour.\newlineAt a certain instant, the surface area is 2424 square meters.\newlineWhat is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 88\newline(B) 152 \frac{15}{2} \newline(C) (15)3 (\sqrt{15})^{3} \newline(D) 58 \frac{5}{8}

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Grade 6right chevron icon
Volume of cubes and rectangular prisms: word problemsright chevron icon

Full solution

Q. The surface area of a cube is increasing at a rate of 1515 square meters per hour.\newlineAt a certain instant, the surface area is 2424 square meters.\newlineWhat is the rate of change of the volume of the cube at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 88\newline(B) 152 \frac{15}{2} \newline(C) (15)3 (\sqrt{15})^{3} \newline(D) 58 \frac{5}{8}
  1. Surface Area Calculation: The surface area of a cube is 66 times one of its side's square, so one side's square is 24 m26\frac{24 \, \text{m}^2}{6}.\newlineCalculate the length of one side.\newlineSide length = 24 m26=4 m2=2 m\sqrt{\frac{24 \, \text{m}^2}{6}} = \sqrt{4 \, \text{m}^2} = 2 \, \text{m}.
  2. Volume Calculation: The volume of a cube is side length cubed, V=s3V = s^3. Calculate the volume with the side length. Volume = 2 m×2 m×2 m=8 m32 \, \text{m} \times 2 \, \text{m} \times 2 \, \text{m} = 8 \, \text{m}^3.
  3. Surface Area Rate of Change: The surface area is increasing at a rate of 15 m2/hour15 \, \text{m}^2/\text{hour}. Since the surface area is 6×side length26 \times \text{side length}^2, d(SA)dt=6×2×side length×d(side length)dt\frac{d(SA)}{dt} = 6 \times 2 \times \text{side length} \times \frac{d(\text{side length})}{dt}.
  4. Side Length Rate of Change: Solve for d(side length)dt\frac{d(\text{side length})}{dt}.15 m2/hour=6×2×2 m×d(side length)dt15 \, \text{m}^2/\text{hour} = 6 \times 2 \times 2 \, \text{m} \times \frac{d(\text{side length})}{dt}.d(side length)dt=15 m2/hour(6×2×2 m)=15 m2/hour24 m=58 m/hour\frac{d(\text{side length})}{dt} = \frac{15 \, \text{m}^2/\text{hour}}{(6 \times 2 \times 2 \, \text{m})} = \frac{15 \, \text{m}^2/\text{hour}}{24 \, \text{m}} = \frac{5}{8} \, \text{m/hour}.
  5. Volume Rate of Change: The rate of change of the volume is given by dVdt=3×side length2×d(side length)dt\frac{dV}{dt} = 3 \times \text{side length}^2 \times \frac{d(\text{side length})}{dt}. Calculate dVdt\frac{dV}{dt} using the side length and d(side length)dt\frac{d(\text{side length})}{dt}. dVdt=3×(2 m)2×(58 m/hour)=3×4 m2×58 m/hour=152 m3/hour.\frac{dV}{dt} = 3 \times (2 \, \text{m})^2 \times \left(\frac{5}{8} \, \text{m/hour}\right) = 3 \times 4 \, \text{m}^2 \times \frac{5}{8} \, \text{m/hour} = \frac{15}{2} \, \text{m}^3/\text{hour}.

More problems from Volume of cubes and rectangular prisms: word problems

Question
Chinedu was given this problem:\newlineA person stands at a distance of 1818 meters east of an intersection and watches a car driving away from the intersection to the north at 33 meters per second. The function y(t) y(t) gives the car's distance, in meters, north of the intersection at time t t . At a certain instant t0 t_{0} , the angle θ(t0) \theta\left(t_{0}\right) between the line of sight from the person to the car and the line of sight from the person to the intersection is 00.77 radians. What is the rate of change of the angle θ(t) \theta(t) at that instant?\newlineWhich equation should Chinedu use to solve the problem?\newlineChoose 11 answer:\newline(A) tan⁡[θ(t)]=y(t)18 \tan [\theta(t)]=\frac{y(t)}{18} \newline(B) θ(t)=18⋅y(t)2 \theta(t)=\frac{18 \cdot y(t)}{2} \newline(C) θ(t)+18+y(t)=180 \theta(t)+18+y(t)=180 \newline(D) [θ(t)]2=182+[y(t)]2 [\theta(t)]^{2}=18^{2}+[y(t)]^{2}
Get tutor helpright-arrow

Posted 9 months ago

Question
Steve was given this problem:\newlineThe radius r(t) r(t) of a circle is increasing at a rate of 33 centimeters per second. At a certain instant t0 t_{0} , the radius is 88 centimeters. What is the rate of change of the area A(t) A(t) of the circle at that instant?\newlineWhich equation should Steve use to solve the problem?\newlineChoose 11 answer:\newline(A) A(t)=2π⋅r(t) A(t)=2 \pi \cdot r(t) \newline(B) A(t)=π[r(t)]2 A(t)=\pi[r(t)]^{2} \newline(C) A(t)=4π[r(t)]2 A(t)=4 \pi[r(t)]^{2} \newline(D) A(t)=43π[r(t)]3 A(t)=\frac{4}{3} \pi[r(t)]^{3}
Get tutor helpright-arrow

Posted 9 months ago

Question
Benjamin invests $400 \$ 400 in a savings account that earns 5% 5 \% interest each year.\newlineWhich expression gives the balance in the account after 33 years?\newlineChoose 11 answer:\newline(A) 400⋅0.05⋅0.05⋅0.05 400 \cdot 0.05 \cdot 0.05 \cdot 0.05 \newline(B) 400+(1+0.05)(1+0.05)(1+0.05) 400+(1+0.05)(1+0.05)(1+0.05) \newline(C) 400+0.05⋅0.05⋅0.05 400+0.05 \cdot 0.05 \cdot 0.05 \newline(D) 400(1+0.05)(1+0.05)(1+0.05) 400(1+0.05)(1+0.05)(1+0.05)
Get tutor helpright-arrow

Posted 9 months ago

Question
Let f(x)=x+1xex+ex f(x)=\frac{x+1}{xe^{x}+e^{x}} when x≠−1 x\neq-1 .\newlinef f is continuous for all real numbers.\newlineFind f(−1) f(-1) .\newlineChoose 11 answer:\newline(A) 1 1 \newline(B) e e \newline(C) 2 2
Get tutor helpright-arrow

Posted 8 months ago

Question
v=3RTM v=\sqrt{\frac{3 R T}{M}} \newlineThe root-mean-square speed is the measure of the speed of particles in a gas. Root-mean-square speed, v v , can be calculated using the equation shown, where M M is the molar mass of a gas, R R is the molar gas constant, and T T is the temperature. Which of the following equations correctly expresses the molar mass of a gas in terms of root-mean-square speed, temperature, and the molar gas constant?\newlineChoose 11 answer:\newline(A) M=(3RTv)2 M=\left(\frac{3 R T}{v}\right)^{2} \newline(B) M=3RT(v)2 M=\frac{3 R T}{(v)^{2}} \newline(C) M=(v)23RT M=\frac{(v)^{2}}{3 R T} \newline(D) M=3RTv M=\frac{\sqrt{3 R T}}{v}
Get tutor helpright-arrow

Posted 7 months ago

Question
f=3π(d2)2h f=3 \pi\left(\frac{d}{2}\right)^{2} h \newlineA board foot is a 11 foot by 11 foot by 11 inch thick volume. The formula estimates the board feet, f f , available in a tree with a diameter of d d inches and a height of h h feet. Which of the following equations correctly gives the diameter in terms of the board feet and the height of the tree?\newlineChoose 11 answer:\newline(A) d=2f3πh d=\sqrt{\frac{2 f}{3 \pi h}} \newline(B) d=2f3πh d=2 \sqrt{\frac{f}{3 \pi h}} \newline(C) d=2f23πh d=\frac{2 f^{2}}{3 \pi h} \newline(D) d=f223πh d=\frac{f^{2} \sqrt{2}}{3 \pi h}
Get tutor helpright-arrow

Posted 7 months ago

Question
The surface area of a cylinder is increasing at a rate of 9π 9 \pi square meters per hour.\newlineThe height of the cylinder is fixed at 33 meters.\newlineAt a certain instant, the surface area is 36π 36 \pi square meters.\newlineWhat is the rate of change of the volume of the cylinder at that instant (in cubic meters per hour)?\newlineChoose 11 answer:\newline(A) 9π 9 \pi \newline(B) 27π 27 \pi \newline(C) π3 \frac{\pi}{3} \newline(D) 12 \frac{1}{2} \newlineThe surface area of a cylinder with base radius r r and height h h is 2πr2+2πrh 2 \pi r^{2}+2 \pi r h .\newlineThe volume of a cylinder with base radius r r and height h h is πr2h \pi r^{2} h .
Get tutor helpright-arrow

Posted 7 months ago

Question
The radius of the base of a cone is increasing at a rate of 1010 meters per second.\newlineThe height of the cone is fixed at 66 meters.\newlineAt a certain instant, the radius is 11 meter.\newlineWhat is the rate of change of the volume of the cone at that instant (in cubic meters per second)?\newlineChoose 11 answer:\newline(A) 20π 20 \pi \newline(B) 40π 40 \pi \newline(C) 400π 400 \pi \newline(D) 200π 200 \pi \newlineThe volume of a cone with radius r r and height h h is πr2h3 \pi r^{2} \frac{h}{3} .
Get tutor helpright-arrow

Posted 7 months ago

Question
The side of the base of a square prism is increasing at a rate of 55 meters per second and the height of the prism is decreasing at a rate of 22 meters per second.\newlineAt a certain instant, the base's side is 66 meters and the height is 77 meters.\newlineWhat is the rate of change of the volume of the prism at that instant (in cubic meters per second)?\newlineChoose 11 answer:\newline(A) 492492\newline(B) −492-492\newline(C) −348-348\newline(D) 348348\newlineThe volume of a square prism with base side s s and height h h is s2h s^{2} h .
Get tutor helpright-arrow

Posted 7 months ago

Question
The side of the base of a square pyramid is increasing at a rate of 66 meters per minute and the height of the pyramid is decreasing at a rate of 11 meter per minute.\newlineAt a certain instant, the base's side is 33 meters and the height is 99 meters.\newlineWhat is the rate of change of the volume of the pyramid at that instant (in cubic meters per minute)?\newlineChoose 11 answer:\newline(A) −111-111\newline(B) 111 \mathbf{1 1 1} \newline(C) −105-105\newline(D) 105105\newlineThe volume of a square pyramid with base side s s and height h h is 13s2h \frac{1}{3} s^{2} h .
Get tutor helpright-arrow

Posted 7 months ago

View all (15)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant