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# Optimization: area of triangle & square (Part 2)

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.B (LO)
,
FUN‑4.B.1 (EK)
,
FUN‑4.C (LO)
,
FUN‑4.C.1 (EK)

## Video transcript

where we left off in the last video we had come up with an expression as a function of X of our combined area based on where we make the cut and now we just need to figure out where this hits a minimum value and to do that we just have to take we just have to take the derivative of this business figure out where our derivative is either undefined or zero and then just make sure that that is a minimum value and then we'll be all set so let me rewrite this or our combined area our combined area as a function of X let me just rewrite this so it's a little bit easier to take the derivative so this is going to be the square root of three times x squared over let's see this is four times nine this is x squared over nine so this is going to be 4 times 9 is 36 and then over here in blue this is going to be plus 100 minus x squared over 16 now let's take the derivative of this so a prime our combined area the derivative of our combined area is a function of X is going to be equal to well the derivative of this with respect to X is just going to be square root of 3x over 18 the derivative of this with respect to X what's the derivative of something squared over 16 with respect to that something so that's going to be that something something to the first power times 2 over 16 which is just over 8 and then time who's doing the chain rule times the derivative of the something with respect to X the derivative 100 minus X with respect to X is just negative 1 so x times negative 1 times so we'll multiply negative 1 right over here and so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus plus let's see I could write this as positive x over 8 so I could write this as 1/8 X right because the negative 1 times the negative x is positive x over 8 and then minus 100 over 8 which is negative 12.5 - 12.5 and we want to figure out an X that minimizes this area so let's this this derivative right over here is defined for any X so we're not going to get our critical point by figuring out where the derivative is undefined but we might get a critical point by setting this derivative equal to zero to figure out what X values make our derivative zero when do we have a zero slope for our original function and then we just have to verify that this is going to be a minimum point if we can find an X that makes this thing equal to zero so let's try to solve for X so if we add 12.5 to both sides we get 12.5 is equal to if you add the X terms you get square root of three over 18 plus one-eighth X to solve for X divide both sides by this business you get X is equal to twelve point five over square root of three over 18 plus one over eight and we are done at x equals this our derivative is equal to zero now we know why I shouldn't say we're done yet we don't know whether this is a minimum point in order to figure out whether this is a minimum point we have to figure out whether our our function is concave upward or concave downward when X is equal to this business and to figure that out let's take the second derivative here so the second derivative so let me rewrite the second derivative of all of this business the second derivative well this is this was the same function as this right over here so let me write it rewrite it so a prime the derivative of our combined area is equal to the square root of three over 18 X plus one-eighth X minus 12.5 the second derivative the second derivative is going to be square root of three over 18 plus one over eight so this thing right over here is greater than zero which means we're concave upward for all X's concave concave upwards which means for all X's we're kind of doing a situation like this so if we find an X where the slope is zero it's going to be over an interval we're concave upwards this is concave upwards for all X so we're going to be we're going to be at a minimum point the slope is right over here this right over here will be a minimum point so once again this is going to be a minimum point now if we actually had 100 meter wire this expression isn't too valuable we would want to get a pretty close approximation in terms of where to actually make the cut an actual decimal number so let's use a calculator to get that so we have 12 point 5 divided by square root of 3 square root of 3 divided by 18 plus 1/8 1/8 gives us and now we deserve our drumroll this is 56 point 5 so this is approximately equal to approximately equal to 56 56 point 5 meters so you make you make this cut 56 roughly 56 point 5 I'll write roughly I'll make it little 56 point 5 meters from the left hand side and you will minimize the combined area of both of these figures