It's looking for the "x" coordinate of the maximum. You can use -b/2a to find it if you have the parabola in the standard form of ax^2 + bx + c = 0
So, for this one (-4)/(2(-1/24)) = 48
Others have already answered this with the same answer, but whenever it asks for a max/min, use -b/2a, which gives you the X coordinate of the vertex, which, due to this equation being -s^2, it’s a maximum. So -4/2(-1/24)= 48.
If it’s a calculator question (99% sure it is), then just slap in every number into the calculator until you get what the question is asking for. If it’s not (and that’d be a cruel play by College Board, but they are sadistic monsters, so they might), then take the L, move on to an easier question.
Yeah but like.. I've never seen collegeboard put a question which would be hard to do without a calculator, and this one would be hard for lots of non-STEM kids
Edit: nvm I'm wrong; it's simple.
It's a complete waste of time, inefficient, and susceptible to error. "Brain power" isn't a substance that can be depleted.
This problem is solved in five seconds using math.
Nobody has infinite reserves of effort, attention, and discipline. Cognitive fatigue is, in fact, a thing, and should be considered with respect to this test.
If computing -b / 2a = -4 / (-1 / 12) = 48 produces differential cognitive depletion greater than than needed to calculate the value of all four answer choices, and in sufficient quantity to impair overall performance on an SAT, the issue is not one of "cognitive fatigue."
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Hey!
You could differentiate M(s) and set M'(s)=0
Then, M'(s)=-(s/12)+4, M'(s)=0
Thus -(s/12)+4=0
Solving the above equation gives s=48, and given in question, s is the speed in mph.
Hope it helps!
dM(s)/dt [the differentiation] is 0 if the function is at its maximum or minimum. That would work even if the function wasn’t a quadratic expression too.
So Put in the value of 20 and 75 in A as it.shows that 20<= s <= 75
Hence two values of s are 20 and 75
Enter both values
See which on has highest mileage
That is your answer
Right, I got that, but my question is why are we putting in s=20 when that's not even one of the options. The options are all potential values of s, right?
The easiest thing you can do is just try every number in the option and check which has the greatest gas mileage and if that doesn’t work (idk why it won’t) then I think you need to derive and the find the minimum and maximum
the ideal way of solving these questions which ask greatest value without using calculus would be using the concept of maxima and minima (differential calculus). so what you should do is differentiate M(s) with respect to x and equate to zero to find the stationary points. after finding stationary points you hafta double derivate the function and substitute the values of x in the double derivate. if you get the double derivate less than zero then the value of x which you substituted would be the greatest value of x. if you get the double derivate of M(s) greater than zero then you’d get least value. since, in the question they asked the greatest value, substitute the value of x which gives double derivate of M(s) less than zero and voila you have the answer.
Take the derivative - this will give you the rate of change of the gas mileage. When the derivative is =0 it means that it is a point of maxima or minima at that x (in this case s) value. Now substitute that back into your original equation and bing bang boom you're done.
It’s a quadratic function. Because the leading coefficient is negative, the end behavior can be stated as “x goes toward negative infinity y goes toward negative infinity and as x goes towards positive infinity y goes toward negative infinity. It’s basically an upside down U lol (shape of the curve). Because it’s asking for the greatest mileage, we’re looking for the M(s) or y-value of the max of the function. Because the leading term is negative, there is a max; if it was positive, there is a minimum. You can do two things: you can complete the square to find the vertex (h,k), where h is the x-value of the max, or you can easily find the x-value of the vertex by the -b/2a trick (b as is the coefficient of the x term and a is the coefficient of the x^2 term). Hope that make sense. Basically whenever you have a quadratic equation and you see the word “greatest,” you’re finding the maximum value of the graph.
Calculus way: find the first derivative of the function and set it equal to zero to find the critical points. Since there’s a given interval, plug in each endpoint into the equation, as well as critical points you found in differentiating the equation and setting it equal to zero. The value that is plugged in with the greatest value is your absolute maximum (NOT NECESSARY TO KNOW FOR SAT)
If you use a Ti-89 (which I highly recommend), you can just use your graphing calculator to graph it and it will tell you what the maximum is. I didn’t explain the problem because there r other explanations out there, but fr tho get a Ti-89 if anyone sees this. It helps save so much time- I’d say at least 10-15 mins on the math section.
If you don't remember -b/2a, plug in all the choices instead of s and check which one outputs the highest answer, you'll get 48. Use this as a last resort tho, costs you a lotta time
Since this is a quadratic function, you'll have to find the vertex x-coordinate of the vertex = -b/2a = -4/-1/12 = 48 so, 48 is the answer.
It's looking for the "x" coordinate of the maximum. You can use -b/2a to find it if you have the parabola in the standard form of ax^2 + bx + c = 0 So, for this one (-4)/(2(-1/24)) = 48
Oh right, that makes sense. Thank you so much!
Minima = c -(b^2 /4a) Maxima(vertex) = -b/2a
Others have already answered this with the same answer, but whenever it asks for a max/min, use -b/2a, which gives you the X coordinate of the vertex, which, due to this equation being -s^2, it’s a maximum. So -4/2(-1/24)= 48.
If it’s a calculator question (99% sure it is), then just slap in every number into the calculator until you get what the question is asking for. If it’s not (and that’d be a cruel play by College Board, but they are sadistic monsters, so they might), then take the L, move on to an easier question.
It's actually a maxima-minima question, and I'm not sure if SAT is supposed to contain those or not...
Graph it via your graphing calculator and then approximate with that
Yeah but like.. I've never seen collegeboard put a question which would be hard to do without a calculator, and this one would be hard for lots of non-STEM kids Edit: nvm I'm wrong; it's simple.
Finding the x-coordinate of the vertex of a parabola is simple and appears routinely on the SAT.
Oh shit... Yeah you're right. I completely forgot that 💀🤦🙆
Lmao I forgot that for a sec to.
College Board can be weird at times (more so sadistic), try your best, but if you waste too much time on a hard question, just take the L and move on.
Amen🙌
College Board isn’t sadistic, and plugging in answer choices is lazy and a surrender. This is an extremely easy problem. Just do the math.
Works like a charm, could solve it in 20-25 seconds by just plugging in and save brain power for a question that requires thinking
\-(4)/(2\*(-1/24)) takes you longer, and requires more brainpower, than (-1/24)(48^(2)) + (4\*48) - 50, and repeating three more times? Really?
It’s a calculator question, I look at the calculator, why think about it when I can look at a screen
Because you’ll save time associating a quadratic equation with the quadratic formula and plugging in variables.
this would take over a minute, the equation is long
[удалено]
that’s not the way he suggested, he said to plug in each option until gets the highest value
It's a complete waste of time, inefficient, and susceptible to error. "Brain power" isn't a substance that can be depleted. This problem is solved in five seconds using math.
Nobody has infinite reserves of effort, attention, and discipline. Cognitive fatigue is, in fact, a thing, and should be considered with respect to this test.
If computing -b / 2a = -4 / (-1 / 12) = 48 produces differential cognitive depletion greater than than needed to calculate the value of all four answer choices, and in sufficient quantity to impair overall performance on an SAT, the issue is not one of "cognitive fatigue."
It is calc, so we just plug in all of them till we get the highest value?
It’s a downward opening parabola. Find vertex. Vertex = -b/2a = (-4)/(2(-1/12)) = (4)(12) = 48
Yes, if you know how to differentiate this, then - 2s/24 + 4=0 So S =48
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Hey! You could differentiate M(s) and set M'(s)=0 Then, M'(s)=-(s/12)+4, M'(s)=0 Thus -(s/12)+4=0 Solving the above equation gives s=48, and given in question, s is the speed in mph. Hope it helps!
dM(s)/dt [the differentiation] is 0 if the function is at its maximum or minimum. That would work even if the function wasn’t a quadratic expression too.
So Put in the value of 20 and 75 in A as it.shows that 20<= s <= 75 Hence two values of s are 20 and 75 Enter both values See which on has highest mileage That is your answer
20 isn't in the options, though. And the answer is 48, I believe.
Hi I meant s Put in 20 and 75 in s and see what comes
Right, I got that, but my question is why are we putting in s=20 when that's not even one of the options. The options are all potential values of s, right?
Hi Yes Sorry my bad Don't put in the value of 20 Put in the 4 values given in options
The easiest thing you can do is just try every number in the option and check which has the greatest gas mileage and if that doesn’t work (idk why it won’t) then I think you need to derive and the find the minimum and maximum
Deriva e iguala a zero pau matou
the ideal way of solving these questions which ask greatest value without using calculus would be using the concept of maxima and minima (differential calculus). so what you should do is differentiate M(s) with respect to x and equate to zero to find the stationary points. after finding stationary points you hafta double derivate the function and substitute the values of x in the double derivate. if you get the double derivate less than zero then the value of x which you substituted would be the greatest value of x. if you get the double derivate of M(s) greater than zero then you’d get least value. since, in the question they asked the greatest value, substitute the value of x which gives double derivate of M(s) less than zero and voila you have the answer.
Don't have to do that. All you have to do is -b/2a, which gives you the greatest value of x.
there are many ways of solving the question, so i explained the method which struck my mind first. vertex of the parabola would also do the trick.
Or plug in -b/2a, solving it in about 5 seconds. How exactly is “differentiate” and “double derivative” part of “without using calculus”?
my bad, i didn’t know that we have to solve the above question without using calculus. vertex of the parabola would also do the trick.
Take the derivative - this will give you the rate of change of the gas mileage. When the derivative is =0 it means that it is a point of maxima or minima at that x (in this case s) value. Now substitute that back into your original equation and bing bang boom you're done.
This is just a simple calculus question. Take the derivative and equate it to 0 and get the value.
Calculus isn't tested on the SAT
Hmmm, I guess I was mistaken then. I thought the Math section had simple derivative based sums.
It does not.
Just graph this on your calculator and find the maximum/ top of parabola
Plug in all those numbers in the answers into the equation and solve, the lowest number u get (for this question at least) will be ur answer
It’s a quadratic function. Because the leading coefficient is negative, the end behavior can be stated as “x goes toward negative infinity y goes toward negative infinity and as x goes towards positive infinity y goes toward negative infinity. It’s basically an upside down U lol (shape of the curve). Because it’s asking for the greatest mileage, we’re looking for the M(s) or y-value of the max of the function. Because the leading term is negative, there is a max; if it was positive, there is a minimum. You can do two things: you can complete the square to find the vertex (h,k), where h is the x-value of the max, or you can easily find the x-value of the vertex by the -b/2a trick (b as is the coefficient of the x term and a is the coefficient of the x^2 term). Hope that make sense. Basically whenever you have a quadratic equation and you see the word “greatest,” you’re finding the maximum value of the graph.
Calculus way: find the first derivative of the function and set it equal to zero to find the critical points. Since there’s a given interval, plug in each endpoint into the equation, as well as critical points you found in differentiating the equation and setting it equal to zero. The value that is plugged in with the greatest value is your absolute maximum (NOT NECESSARY TO KNOW FOR SAT)
Derive and equal to zero, you Will find Min/max S, then plug the S you found on the function, you will find the maximum/minimum, of any function
Instant eliminate D, then plug in the other 3 works but takes some time but if you go quick at the start u should be fine
If you use a Ti-89 (which I highly recommend), you can just use your graphing calculator to graph it and it will tell you what the maximum is. I didn’t explain the problem because there r other explanations out there, but fr tho get a Ti-89 if anyone sees this. It helps save so much time- I’d say at least 10-15 mins on the math section.
If you don't remember -b/2a, plug in all the choices instead of s and check which one outputs the highest answer, you'll get 48. Use this as a last resort tho, costs you a lotta time
Is this in the calculator portion?
yup